The Real And Complex Number Systems
The Real And Complex Number Systems
The Real And Complex Number Systems
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<strong>The</strong> <strong>Real</strong> <strong>And</strong> <strong>Complex</strong> <strong>Number</strong> <strong>Systems</strong><br />
Integers<br />
1.1 Prove that there is no largest prime.<br />
Proof: Suppose p is the largest prime. <strong>The</strong>n p! + 1 is NOT a prime. So,<br />
there exists a prime q such that<br />
q |p! + 1 ⇒ q |1<br />
which is impossible. So, there is no largest prime.<br />
Remark: <strong>The</strong>re are many and many proofs about it. <strong>The</strong> proof that we<br />
give comes from Archimedes 287-212 B. C. In addition, Euler Leonhard<br />
(1707-1783) find another method to show it. <strong>The</strong> method is important since<br />
it develops to study the theory of numbers by analytic method. <strong>The</strong> reader<br />
can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s by<br />
Loo-Keng Hua, pp 91-93. (Chinese Version)<br />
1.2 If n is a positive integer, prove the algebraic identity<br />
∑n−1<br />
a n − b n = (a − b) a k b n−1−k<br />
k=0<br />
Proof: It suffices to show that<br />
∑n−1<br />
x n − 1 = (x − 1) x k .<br />
k=0<br />
1
Consider the right hand side, we have<br />
∑n−1<br />
∑n−1<br />
∑n−1<br />
(x − 1) x k = x k+1 −<br />
k=0<br />
=<br />
k=0<br />
n∑<br />
x k −<br />
k=1<br />
= x n − 1.<br />
k=0<br />
∑n−1<br />
x k<br />
k=0<br />
x k<br />
1.3 If 2 n − 1 is a prime, prove that n is prime. A prime of the form<br />
2 p − 1, where p is prime, is called a Mersenne prime.<br />
Proof: If n is not a prime, then say n = ab, where a > 1 and b > 1. So,<br />
we have<br />
∑b−1<br />
2 ab − 1 = (2 a − 1) (2 a ) k<br />
which is not a prime by Exercise 1.2. So, n must be a prime.<br />
Remark: <strong>The</strong> study of Mersenne prime is important; it is related<br />
with so called Perfect number. In addition, there are some OPEN problem<br />
about it. For example, is there infinitely many Mersenne nembers<br />
<strong>The</strong> reader can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory<br />
Of <strong>Number</strong>s by Loo-Keng Hua, pp 13-15. (Chinese Version)<br />
1.4 If 2 n + 1 is a prime, prove that n is a power of 2. A prime of the<br />
form 2 2m + 1 is called a Fermat prime. Hint. Use exercise 1.2.<br />
Proof: If n is a not a power of 2, say n = ab, where b is an odd integer.<br />
So,<br />
2 a + 1 ∣ ∣ 2 ab + 1<br />
and 2 a + 1 < 2 ab + 1. It implies that 2 n + 1 is not a prime. So, n must be a<br />
power of 2.<br />
k=0<br />
Remark: (1) In the proof, we use the identity<br />
2n−2<br />
∑<br />
x 2n−1 + 1 = (x + 1) (−1) k x k .<br />
k=0<br />
2
Proof: Consider<br />
2n−2<br />
∑<br />
(x + 1) (−1) k x k =<br />
k=0<br />
=<br />
2n−2<br />
∑<br />
k=0<br />
2n−1<br />
(−1) k x k+1 +<br />
2n−2<br />
∑<br />
k=0<br />
2n−2<br />
(−1) k x k<br />
∑<br />
∑<br />
(−1) k+1 x k + (−1) k x k<br />
k=1<br />
= x 2n+1 + 1.<br />
k=0<br />
(2) <strong>The</strong> study of Fermat number is important; for the details the reader<br />
can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s by<br />
Loo-Keng Hua, pp 15. (Chinese Version)<br />
1.5 <strong>The</strong> Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, ... are defined by the recursion<br />
formula x n+1 = x n + x n−1 , with x 1 = x 2 = 1. Prove that (x n , x n+1 ) = 1<br />
and that x n = (a n − b n ) / (a − b) , where a and b are the roots of the quadratic<br />
equation x 2 − x − 1 = 0.<br />
So,<br />
Proof: Let d = g.c.d. (x n , x n+1 ) , then<br />
Continue the process, we finally have<br />
So, d = 1 since d is positive.<br />
Observe that<br />
and thus we consider<br />
i.e., consider<br />
If we let<br />
d |x n and d |x n+1 = x n + x n−1 .<br />
d |x n−1 .<br />
d |1 .<br />
x n+1 = x n + x n−1 ,<br />
x n+1 = x n + x n−1 ,<br />
x 2 = x + 1 with two roots, a and b.<br />
F n = (a n − b n ) / (a − b) ,<br />
3
then it is clear that<br />
So, F n = x n for all n.<br />
F 1 = 1, F 2 = 1, and F n+1 = F n + F n−1 for n > 1.<br />
Remark: <strong>The</strong> study of the Fibonacci numbers is important; the reader<br />
can see the book, Fibonacci and Lucas <strong>Number</strong>s with Applications<br />
by Koshy and Thomas.<br />
1.6 Prove that every nonempty set of positive integers contains a smallest<br />
member. This is called the well–ordering Principle.<br />
Proof: Given (φ ≠) S (⊆ N) , we prove that if S contains an integer<br />
k, then S contains the smallest member. We prove it by Mathematical<br />
Induction of second form as follows.<br />
As k = 1, it trivially holds. Assume that as k = 1, 2, ..., m holds, consider<br />
as k = m + 1 as follows. In order to show it, we consider two cases.<br />
(1) If there is a member s ∈ S such that s < m + 1, then by Induction<br />
hypothesis, we have proved it.<br />
(2) If every s ∈ S, s ≥ m + 1, then m + 1 is the smallest member.<br />
Hence, by Mathematical Induction, we complete it.<br />
Remark: We give a fundamental result to help the reader get more. We<br />
will prove the followings are equivalent:<br />
(A. Well–ordering Principle) every nonempty set of positive integers<br />
contains a smallest member.<br />
(B. Mathematical Induction of first form) Suppose that S (⊆ N) ,<br />
if S satisfies that<br />
<strong>The</strong>n S = N.<br />
(1). 1 in S<br />
(2). As k ∈ S, then k + 1 ∈ S.<br />
(C. Mathematical Induction of second form) Suppose that S (⊆ N) ,<br />
if S satisfies that<br />
(1). 1 in S<br />
(2). As 1, ..., k ∈ S, then k + 1 ∈ S.<br />
4
<strong>The</strong>n S = N.<br />
Proof: (A ⇒ B): If S ≠ N, then N − S ≠ φ. So, by (A), there exists<br />
the smallest integer w such that w ∈ N − S. Note that w > 1 by (1), so we<br />
consider w − 1 as follows.<br />
Since w − 1 /∈ N − S, we know that w − 1 ∈ S. By (2), we know that<br />
w ∈ S which contadicts to w ∈ N − S. Hence, S = N.<br />
(B ⇒ C): It is obvious.<br />
(C ⇒ A): We have proved it by this exercise.<br />
Rational and irrational numbers<br />
1.7 Find the rational number whose decimal expansion is 0.3344444444....<br />
Proof: Let x = 0.3344444444..., then<br />
x = 3 10 + 3<br />
10 + 4<br />
2 10 + ... + 4 + .., where n ≥ 3<br />
3 10n = 33<br />
10 + 4 (<br />
1 + 1 2 10 3 10 + ... + 1 )<br />
10 + .. n<br />
= 33<br />
10 + 4 ( 1<br />
2 10 3 1 − 1<br />
10<br />
= 33<br />
10 + 4<br />
2 900<br />
= 301<br />
900 .<br />
)<br />
1.8 Prove that the decimal expansion of x will end in zeros (or in nines)<br />
if, and only if, x is a rational number whose denominator is of the form 2 n 5 m ,<br />
where m and n are nonnegative integers.<br />
Proof: (⇐)Suppose that x =<br />
k , if n ≥ m, we have<br />
2 n 5 m<br />
k5 n−m<br />
2 n 5 n = 5n−m k<br />
10 n .<br />
So, the decimal expansion of x will end in zeros. Similarly for m ≥ n.<br />
(⇒)Suppose that the decimal expansion of x will end in zeros (or in<br />
nines).<br />
5
For case x = a 0 .a 1 a 2 · · · a n . <strong>The</strong>n<br />
∑ n<br />
k=0<br />
x =<br />
10n−k a k<br />
=<br />
10 n<br />
∑ n<br />
k=0 10n−k a k<br />
2 n 5 n .<br />
For case x = a 0 .a 1 a 2 · · · a n 999999 · · · . <strong>The</strong>n<br />
∑ n<br />
k=0<br />
x =<br />
10n−k a k<br />
+ 9<br />
9<br />
+ ... + + ...<br />
2 n 5 n 10n+1 10n+m ∑ n<br />
k=0<br />
=<br />
10n−k a k<br />
+ 9 ∑ ∞<br />
10 −j<br />
2 n 5 n 10 n+1 j=0<br />
∑ n<br />
k=0<br />
=<br />
10n−k a k<br />
+ 1<br />
2 n 5 n 10 n<br />
= 1 + ∑ n<br />
k=0 10n−k a k<br />
.<br />
2 n 5 n<br />
So, in both case, we prove that x is a rational number whose denominator is<br />
of the form 2 n 5 m , where m and n are nonnegative integers.<br />
1.9 Prove that √ 2 + √ 3 is irrational.<br />
Proof: If √ 2 + √ 3 is rational, then consider<br />
(√<br />
3 +<br />
√<br />
2<br />
) (√<br />
3 −<br />
√<br />
2<br />
)<br />
= 1<br />
which implies that √ 3 − √ 2 is rational. Hence, √ 3 would be rational. It is<br />
impossible. So, √ 2 + √ 3 is irrational.<br />
Remark: (1) √ p is an irrational if p is a prime.<br />
Proof: If √ p ∈ Q, write √ p = a , where g.c.d. (a, b) = 1. <strong>The</strong>n<br />
b<br />
Write a = pq. So,<br />
By (*) and (*’), we get<br />
b 2 p = a 2 ⇒ p ∣ ∣ a 2 ⇒ p |a (*)<br />
b 2 p = p 2 q 2 ⇒ b 2 = pq 2 ⇒ p ∣ ∣ b 2 ⇒ p |b . (*’)<br />
p |g.c.d. (a, b) = 1<br />
which implies that p = 1, a contradiction. So, √ p is an irrational if p is a<br />
prime.<br />
6
Note: <strong>The</strong>re are many and many methods to prove it. For example, the<br />
reader can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s<br />
by Loo-Keng Hua, pp 19-21. (Chinese Version)<br />
(2) Suppose a, b ∈ N. Prove that √ a+ √ b is rational if and only if, a = k 2<br />
and b = h 2 for some h, k ∈ N.<br />
Proof: (⇐) It is clear.<br />
(⇒) Consider<br />
( √a +<br />
√<br />
b<br />
) ( √a −<br />
√<br />
b<br />
)<br />
= a 2 − b 2 ,<br />
then √ a ∈ Q and √ b ∈ Q. <strong>The</strong>n it is clear that a = h 2 and b = h 2 for some<br />
h, k ∈ N.<br />
1.10 If a, b, c, d are rational and if x is irrational, prove that (ax + b) / (cx + d)<br />
is usually irrational. When do exceptions occur<br />
Proof: We claim that (ax + b) / (cx + d) is rational if and only if ad = bc.<br />
(⇒)If (ax + b) / (cx + d) is rational, say (ax + b) / (cx + d) = q/p. We<br />
consider two cases as follows.<br />
(i) If q = 0, then ax + b = 0. If a ≠ 0, then x would be rational. So, a = 0<br />
and b = 0. Hence, we have<br />
ad = 0 = bc.<br />
(ii) If q ≠ 0, then (pa − qc) x+(pb − qd) = 0. If pa−qc ≠ 0, then x would<br />
be rational. So, pa − qc = 0 and pb − qd = 0. It implies that<br />
qcb = qad ⇒ ad = bc.<br />
(⇐)Suppose ad = bc. If a = 0, then b = 0 or c = 0. So,<br />
If a ≠ 0, then d = bc/a. So,<br />
{<br />
ax + b 0 if a = 0 and b = 0<br />
cx + d = b<br />
.<br />
if a = 0 and c = 0<br />
ax + b<br />
cx + d =<br />
d<br />
ax + b<br />
cx + bc/a<br />
=<br />
a (ax + b)<br />
c (ax + b) = a c .<br />
Hence, we proved that if ad = bc, then (ax + b) / (cx + d) is rational.<br />
7
1.11 Given any real x > 0, prove that there is an irrational number<br />
between 0 and x.<br />
Proof: If x ∈ Q c , we choose y = x/2 ∈ Q c . <strong>The</strong>n 0 < y < x. If x ∈ Q,<br />
we choose y = x/ √ 2 ∈ Q, then 0 < y < x.<br />
Remark: (1) <strong>The</strong>re are many and many proofs about it. We may prove<br />
it by the concept of Perfect set. <strong>The</strong> reader can see the book, Principles<br />
of Mathematical Analysis written by Walter Rudin, <strong>The</strong>orem 2.43,<br />
pp 41. Also see the textbook, Exercise 3.25.<br />
(2) Given a and b ∈ R with a < b, there exists r ∈ Q c , and q ∈ Q such<br />
that a < r < b and a < q < b.<br />
Proof: We show it by considering four cases. (i) a ∈ Q, b ∈ Q. (ii)<br />
a ∈ Q, b ∈ Q c . (iii) a ∈ Q c , b ∈ Q. (iv) a ∈ Q c , b ∈ Q c .(<br />
)<br />
1 − √ 1<br />
2<br />
b.<br />
(i) (a ∈ Q, b ∈ Q) Choose q = a+b<br />
2<br />
and r = √ 1<br />
2<br />
a +<br />
(ii) (a ∈ Q, b ∈ Q c ) Choose r = a+b<br />
2<br />
and let c = 1<br />
< b−a, then a+c := q.<br />
2 n<br />
(iii) (a ∈ Q c , b ∈ Q) Similarly for (iii).<br />
(iv) (a ∈ Q c , b ∈ Q c ) It suffices to show that there exists a rational<br />
number q ∈ (a, b) by (ii). Write<br />
Choose n large enough so that<br />
b = b 0 .b 1 b 2 · · · b n · ··<br />
a < q = b 0 .b 1 b 2 · · · b n < b.<br />
(It works since b − q = 0.000..000b n+1 ... ≤ 1<br />
10 n )<br />
1.12 If a/b < c/d with b > 0, d > 0, prove that (a + c) / (b + d) lies<br />
bwtween the two fractions a/b and c/d<br />
Proof: It only needs to conisder the substraction. So, we omit it.<br />
Remark: <strong>The</strong> result of this exercise is often used, so we suggest the<br />
reader keep it in mind.<br />
1.13 Let a and b be positive integers. Prove that √ 2 always lies between<br />
the two fractions a/b and (a + 2b) / (a + b) . Which fraction is closer to √ 2<br />
Proof: Suppose a/b ≤ √ 2, then a ≤ √ 2b. So,<br />
a + 2b<br />
a + b − √ (√ ) (√ )<br />
2 − 1 2b − a<br />
2 =<br />
≥ 0.<br />
a + b<br />
8
In addition,<br />
(√ a<br />
)<br />
2 − −<br />
b<br />
( a + 2b<br />
a + b − √ )<br />
2 = 2 √ 2 −<br />
( a<br />
b + a + 2b )<br />
a + b<br />
= 2 √ 2 − a2 + 2ab + 2b 2<br />
ab + b 2<br />
1<br />
[(<br />
= 2 √ )<br />
2 − 2 ab +<br />
ab + b 2<br />
≥<br />
= 0.<br />
1<br />
ab + b 2 [ (<br />
2 √ 2 − 2<br />
)<br />
a√ a + 2<br />
(<br />
2 √ ) ]<br />
2 − 2 b 2 − a 2<br />
(<br />
2 √ ) ( ) ] 2<br />
a<br />
2 − 2 √2 − a 2<br />
So, a+2b is closer to √ 2.<br />
a+b<br />
Similarly, we also have if a/b > √ 2, then a+2b<br />
a+b<br />
to √ 2 in this case.<br />
Remark: Note that<br />
a<br />
b < √ 2 < a + 2b<br />
a + b < 2b<br />
a<br />
< √ 2. Also, a+2b<br />
a+b<br />
by Exercise 12 and 13.<br />
is closer<br />
<strong>And</strong> we know that a+2b is closer to √ 2. We can use it to approximate √ 2.<br />
a+b<br />
Similarly for the case<br />
2b<br />
a < a + 2b<br />
a + b < √ 2 < a b .<br />
1.14 Prove that √ n − 1 + √ n + 1 is irrational for every integer n ≥ 1.<br />
Proof: Suppose that √ n − 1 + √ n + 1 is rational, and thus consider<br />
( √n<br />
+ 1 +<br />
√<br />
n − 1<br />
) ( √n<br />
+ 1 −<br />
√<br />
n − 1<br />
)<br />
= 2<br />
which implies that √ n + 1 − √ n − 1 is rational. Hence, √ n + 1 and √ n − 1<br />
are rational. So, n − 1 = k 2 and n + 1 = h 2 , where k and h are positive<br />
integer. It implies that<br />
h = 3 2 and k = 1 2<br />
which is absurb. So, √ n − 1 + √ n + 1 is irrational for every integer n ≥ 1.<br />
9
1.15 Given a real x and an integer N > 1, prove that there exist integers<br />
h and k with 0 < k ≤ N such that |kx − h| < 1/N. Hint. Consider the N +1<br />
numbers tx − [tx] for t = 0, 1, 2, ..., N and show that some pair differs by at<br />
most 1/N.<br />
Proof: Given N > 1, and thus consider tx − [tx] for t = 0, 1, 2, ..., N as<br />
follows. Since<br />
0 ≤ tx − [tx] := a t < 1,<br />
so there exists two numbers a i and a j where i ≠ j such that<br />
|a i − a j | < 1 N ⇒ |(i − j) x − p| < 1 , where p = [jx] − [ix] .<br />
N<br />
Hence, there exist integers h and k with 0 < k ≤ N such that |kx − h| < 1/N.<br />
1.16 If x is irrational prove that there are infinitely many rational numbers<br />
h/k with k > 0 such that |x − h/k| < 1/k 2 . Hint. Assume there are<br />
only a finite number h 1 /k 1 , ..., h r /k r and obtain a contradiction by applying<br />
Exercise 1.15 with N > 1/δ, where δ is the smallest of the numbers<br />
|x − h i /k i | .<br />
Proof: Assume there are only a finite number h 1 /k 1 , ..., h r /k r and let<br />
δ = min r i=1 |x − h i /k i | > 0 since x is irrational. Choose N > 1/δ, then by<br />
Exercise 1.15, we have<br />
∣<br />
1 ∣∣∣<br />
N < δ ≤ x − h k ∣ < 1<br />
kN<br />
which implies that<br />
1<br />
N < 1<br />
kN<br />
which is impossible. So, there are infinitely many rational numbers h/k with<br />
k > 0 such that |x − h/k| < 1/k 2 .<br />
Remark: (1) <strong>The</strong>re is another proof by continued fractions. <strong>The</strong><br />
reader can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s<br />
by Loo-Keng Hua, pp 270. (Chinese Version)<br />
(2) <strong>The</strong> exercise is useful to help us show the following lemma. {ar + b : a ∈ Z, b ∈ Z} ,<br />
where r ∈ Q c is dense in R. It is equivalent to {ar : a ∈ Z} , where r ∈ Q c is<br />
dense in [0, 1] modulus 1.<br />
10
Proof: Say {ar + b : a ∈ Z, b ∈ Z} = S, and since r ∈ Q c , then by Exercise<br />
1.16, there are infinitely many rational numbers h/k with k > 0 such<br />
that |kr − h| < 1 . Consider (x − δ, x + δ) := I, where δ > 0, and thus choosing<br />
k 0 large enough so that 1/k 0 < δ. Define L = |k 0 r − h 0 | , then we have<br />
k<br />
sL ∈ I for some s ∈ Z. So, sL = (±) [(sk 0 ) r − (sh 0 )] ∈ S. That is, we have<br />
proved that S is dense in R.<br />
1.17 Let x be a positive rational number of the form<br />
x =<br />
n∑<br />
k=1<br />
where each a k is nonnegative integer with a k ≤ k − 1 for k ≥ 2 and a n > 0.<br />
Let [x] denote the largest integer in x. Prove that a 1 = [x] , that a k =<br />
[k!x] − k [(k − 1)!x] for k = 2, ..., n, and that n is the smallest integer such<br />
that n!x is an integer. Conversely, show that every positive rational number<br />
x can be expressed in this form in one and only one way.<br />
and<br />
Proof: (⇒)First,<br />
[ ]<br />
n∑ a k<br />
[x] = a 1 +<br />
k!<br />
k=2<br />
[ n∑<br />
]<br />
a k<br />
= a 1 + since a 1 ∈ N<br />
k!<br />
k=2<br />
n∑ a k<br />
n∑<br />
= a 1 since<br />
k! ≤ k − 1<br />
k!<br />
k=2<br />
k=2<br />
Second, fixed k and consider<br />
k!x = k!<br />
n∑<br />
j=1<br />
(k − 1)!x = (k − 1)!<br />
=<br />
k−1<br />
a j<br />
j! = k! ∑<br />
n∑<br />
j=1<br />
j=1<br />
a k<br />
k! ,<br />
n∑<br />
k=2<br />
1<br />
(k − 1)! − 1 k! = 1 − 1 n! < 1.<br />
a j<br />
j! + a k + k!<br />
k−1<br />
a j<br />
j! = (k − 1)! ∑<br />
j=1<br />
n∑<br />
j=k+1<br />
a j<br />
j!<br />
a j<br />
n∑<br />
j! + (k − 1)!<br />
j=k<br />
a j<br />
j! .<br />
11
So,<br />
and<br />
[<br />
∑k−1<br />
[k!x] = k!<br />
j=1<br />
∑k−1<br />
= k!<br />
j=1<br />
a j<br />
j! + a k + k!<br />
a j<br />
j! + a k since k!<br />
[<br />
∑k−1<br />
k [(k − 1)!x] = k (k − 1)!<br />
j=1<br />
∑k−1<br />
= k (k − 1)!<br />
∑k−1<br />
= k!<br />
j=1<br />
a j<br />
j!<br />
j=1<br />
n∑<br />
j=k+1<br />
n∑<br />
a j<br />
j!<br />
j=k+1<br />
]<br />
a j<br />
j! < 1<br />
a j<br />
n∑<br />
j! + (k − 1)!<br />
j=k<br />
]<br />
a j<br />
j! .<br />
a j<br />
n∑<br />
j! since (k − 1)!<br />
j=k<br />
a j<br />
j! < 1<br />
which implies that<br />
a k = [k!x] − k [(k − 1)!x] for k = 2, ..., n.<br />
Last, in order to show that n is the smallest integer such that n!x is an<br />
integer. It is clear that<br />
n∑ a k<br />
n!x = n!<br />
k! ∈ Z.<br />
In addition,<br />
So, we have proved it.<br />
k=1<br />
(n − 1)!x = (n − 1)!<br />
n∑<br />
k=1<br />
n−1<br />
∑<br />
= (n − 1)!<br />
k=1<br />
/∈ Z since a n<br />
n<br />
12<br />
a k<br />
k!<br />
a k<br />
k! + a n<br />
n<br />
/∈ Z.
(⇐)It is clear since every a n is uniquely deermined.<br />
Upper bounds<br />
1.18 Show that the sup and the inf of a set are uniquely determined whenever<br />
they exists.<br />
Proof: Given a nonempty set S (⊆ R) , and assume sup S = a and<br />
sup S = b, we show a = b as follows. Suppose that a > b, and thus choose<br />
ε = a−b , then there exists a x ∈ S such that<br />
2<br />
which implies that<br />
b < a + b<br />
2<br />
= a − ε < x < a<br />
b < x<br />
which contradicts to b = sup S. Similarly for a < b. Hence, a = b.<br />
1.19 Find the sup and inf of each of the following sets of real numbers:<br />
(a) All numbers of the form 2 −p + 3 −q + 5 −r , where p, q, and r take on all<br />
positive integer values.<br />
Proof: Define S = {2 −p + 3 −q + 5 −r : p, q, r ∈ N}. <strong>The</strong>n it is clear that<br />
sup S = 1 + 1 + 1 , and inf S = 0.<br />
2 3 5<br />
(b) S = {x : 3x 2 − 10x + 3 < 0}<br />
Proof: Since 3x 2 − 10x + 3 = (x − 3) (3x − 1) , we know that S = ( 1<br />
3 , 3) .<br />
Hence, sup S = 3 and inf S = 1 3 .<br />
(c) S = {x : (x − a) (x − b) (x − c) (x − d) < 0} , where a < b < c < d.<br />
Proof: It is clear that S = (a, b)∪(c, d) . Hence, sup S = d and inf S = a.<br />
1.20 Prove the comparison property for suprema (<strong>The</strong>orem 1.16)<br />
Proof: Since s ≤ t for every s ∈ S and t ∈ T, fixed t 0 ∈ T, then s ≤ t 0<br />
for all s ∈ S. Hence, by Axiom 10, we know that sup S exists. In addition,<br />
it is clear sup S ≤ sup T.<br />
Remark: <strong>The</strong>re is a useful result, we write it as a reference. Let S and T<br />
be two nonempty subsets of R. If S ⊆ T and sup T exists, then sup S exists<br />
and sup S ≤ sup T.<br />
13
Proof: Since sup T exists and S ⊆ T, we know that for every s ∈ S, we<br />
have<br />
s ≤ sup T.<br />
Hence, by Axiom 10, we have proved the existence of sup S. In addition,<br />
sup S ≤ sup T is trivial.<br />
1.21 Let A and B be two sets of positive numbers bounded above, and<br />
let a = sup A, b = sup B. Let C be the set of all products of the form xy,<br />
where x ∈ A and y ∈ B. Prove that ab = sup C.<br />
Proof: Given ε > 0, we want to find an element c ∈ C such that ab−ε <<br />
c. If we can show this, we have proved that sup C exists and equals ab.<br />
Since sup A = a > 0 and sup B = b > 0, we can choose n large enough<br />
such that a − ε/n > 0, b − ε/n > 0, and n > a + b. So, for this ε ′ = ε/n,<br />
there exists a ′ ∈ A and b ′ ∈ B such that<br />
which implies that<br />
a − ε ′ < a ′ and b − ε ′ < b ′<br />
ab − ε ′ (a + b − ε ′ ) < a ′ b ′ since a − ε ′ > 0 and b − ε ′ > 0<br />
which implies that<br />
which implies that<br />
ab − ε n (a + b) < a′ b ′ := c<br />
ab − ε < c.<br />
1.22 Given x > 0, and an integer k ≥ 2. Let a 0 denote the largest integer<br />
≤ x and, assumeing that a 0 , a 1 , ..., a n−1 have been defined, let a n denote the<br />
largest integer such that<br />
a 0 + a 1<br />
k + a 2<br />
k 2 + ... + a n<br />
k n ≤ x.<br />
Note: When k = 10 the integers a 0 , a 1 , ... are the digits in a decimal<br />
representation of x. For general k they provide a representation in<br />
the scale of k.<br />
(a) Prove that 0 ≤ a i ≤ k − 1 for each i = 1, 2, ...<br />
14
Proof: Choose a 0 = [x], and thus consider<br />
[kx − ka 0 ] := a 1<br />
then<br />
0 ≤ k (x − a 0 ) < k ⇒ 0 ≤ a 1 ≤ k − 1<br />
and<br />
a 0 + a 1<br />
k ≤ x ≤ a 0 + a 1<br />
k + 1 k .<br />
Continue the process, we then have<br />
0 ≤ a i ≤ k − 1 for each i = 1, 2, ...<br />
and<br />
a 0 + a 1<br />
k + a 2<br />
k 2 + ... + a n<br />
k n ≤ x < a 0 + a 1<br />
k + a 2<br />
k 2 + ... + a n<br />
k n + 1<br />
k n . (*)<br />
(b) Let r n = a 0 + a 1 k −1 + a 2 k −2 + ... + a n k −n and show that x is the sup<br />
of the set of rational numbers r 1 , r 2 , ...<br />
Proof: It is clear by (a)-(*).<br />
Inequality<br />
1.23 Prove Lagrange’s identity for real numbers:<br />
( n∑<br />
) 2 ( n∑<br />
) ( n∑<br />
)<br />
a k b k = a 2 k b 2 k −<br />
∑<br />
(a k b j − a j b k ) 2 .<br />
k=1<br />
k=1 k=1 1≤k
and<br />
( n∑<br />
) ( n∑<br />
)<br />
a k b k a k b k = ∑<br />
a k b k a j b j =<br />
k=1<br />
k=1<br />
1≤k,j≤n<br />
So,<br />
( n∑<br />
k=1<br />
a k b k<br />
) 2<br />
=<br />
=<br />
=<br />
( n∑<br />
k=1<br />
( n∑<br />
k=1<br />
( n∑<br />
k=1<br />
a 2 k<br />
a 2 k<br />
a 2 k<br />
) ( n∑<br />
k=1<br />
) ( n∑<br />
k=1<br />
) ( n∑<br />
k=1<br />
b 2 k<br />
b 2 k<br />
b 2 k<br />
)<br />
n∑<br />
k=1<br />
a 2 kb 2 k + ∑ k≠j<br />
+ ∑ a k b k a j b j − ∑ a 2 kb 2 j<br />
k≠j<br />
k≠j<br />
)<br />
∑<br />
+ 2 a k b k a j b j −<br />
∑<br />
)<br />
−<br />
1≤k| ≤ ‖x‖ ‖y‖ by Remark (1).<br />
1.24 Prove that for arbitrary real a k , b k , c k we have<br />
( n∑<br />
) 4 ( n∑<br />
) ( n∑<br />
) 2 ( n∑<br />
)<br />
a k b k c k ≤ a 4 k b 2 k c 4 k .<br />
k=1<br />
k=1 k=1 k=1<br />
16
Proof: Use Cauchy-Schwarz inequality twice, we then have<br />
( n∑<br />
) ⎡ 4 ( n∑<br />
) ⎤ 2<br />
2<br />
a k b k c k = ⎣ a k b k c k<br />
⎦<br />
k=1<br />
k=1<br />
( n∑<br />
) 2 ( n∑<br />
) 2<br />
≤ a 2 kc 2 k b 2 k<br />
k=1<br />
k=1<br />
( n∑<br />
) 2 ( n∑<br />
) ( n∑<br />
≤ a 4 k c 4 k<br />
k=1<br />
k=1 k=1<br />
( n∑<br />
) ( n∑<br />
) 2 ( n∑<br />
= a 4 k b 2 k<br />
k=1 k=1 k=1<br />
b 2 k<br />
c 4 k<br />
) 2<br />
)<br />
.<br />
1.25 Prove that Minkowski’s inequality:<br />
( n∑<br />
k=1<br />
(a k + b k ) 2 ) 1/2<br />
≤<br />
a 2 k) 1/2<br />
+<br />
b 2 k) 1/2<br />
.<br />
( n∑<br />
k=1<br />
( n∑<br />
k=1<br />
This is the triangle inequality ‖a + b‖ ≤ ‖a‖+‖b‖ for n−dimensional vectors,<br />
where a = (a 1 , ..., a n ) , b = (b 1 , ..., b n ) and<br />
‖a‖ =<br />
( n∑<br />
k=1<br />
a 2 k) 1/2<br />
.<br />
Proof: Consider<br />
n∑<br />
n∑ n∑<br />
n∑<br />
(a k + b k ) 2 = a 2 k + b 2 k + 2 a k b k<br />
k=1<br />
k=1<br />
k=1<br />
k=1<br />
(<br />
n∑ n∑<br />
n∑<br />
) 1/2 ( n∑ 1/2<br />
≤ a 2 k + b 2 k + 2 a 2 k bk) 2 by Cauchy-Schwarz inequality<br />
k=1 k=1<br />
k=1<br />
k=1<br />
⎡<br />
1/2 ) ⎤ 1/2<br />
2<br />
= ⎣ ak) 2 +<br />
⎦ .<br />
( n∑<br />
( n∑<br />
k=1<br />
k=1<br />
b 2 k<br />
17
So,<br />
( n∑<br />
k=1<br />
(a k + b k ) 2 ) 1/2<br />
≤<br />
a 2 k) 1/2<br />
+<br />
b 2 k) 1/2<br />
.<br />
( n∑<br />
k=1<br />
( n∑<br />
k=1<br />
1.26 If a 1 ≥ ... ≥ a n and b 1 ≥ ... ≥ b n , prove that<br />
( n∑<br />
) ( n∑<br />
) ( n∑<br />
)<br />
a k b k ≤ n a k b k .<br />
k=1 k=1<br />
k=1<br />
Hint. ∑ 1≤j≤k≤n (a k − a j ) (b k − b j ) ≥ 0.<br />
Proof: Consider<br />
0 ≤ ∑<br />
(a k − a j ) (b k − b j ) = ∑<br />
1≤j≤k≤n<br />
which implies that<br />
Since<br />
∑<br />
1≤j≤k≤n<br />
∑<br />
1≤j≤k≤n<br />
a k b j + a j b k =<br />
1≤j≤k≤n<br />
a k b j + a j b k ≤<br />
=<br />
=<br />
∑<br />
1≤j
In addition,<br />
∑<br />
a k b k + a j b j<br />
1≤j≤k≤n<br />
=<br />
= n<br />
n∑<br />
a k b k + na 1 b 1 +<br />
k=1<br />
n∑<br />
a k b k + (n − 1) a 2 b 2 + ... +<br />
k=2<br />
n∑<br />
a k b k + a 1 b 1 + a 2 b 2 + ... + a n b n<br />
k=1<br />
= (n + 1)<br />
n∑<br />
a k b k<br />
k=1<br />
which implies that, by (**),<br />
( n∑<br />
) ( n∑<br />
) ( n∑<br />
)<br />
a k b k ≤ n a k b k .<br />
k=1 k=1<br />
k=1<br />
n∑<br />
k=n−1<br />
a k b k + 2a n−1 b n−1 + ∑ k=n<br />
a k b k<br />
<strong>Complex</strong> numbers<br />
1.27 Express the following complex numbers in the form a + bi.<br />
(a) (1 + i) 3<br />
Solution: (1 + i) 3 = 1 + 3i + 3i 2 + i 3 = 1 + 3i − 3 − i = −2 + 2i.<br />
(b) (2 + 3i) / (3 − 4i)<br />
Solution: 2+3i<br />
3−4i = (2+3i)(3+4i)<br />
(3−4i)(3+4i) = −6+17i<br />
25<br />
= −6<br />
25 + 17<br />
25 i.<br />
(c) i 5 + i 16<br />
Solution: i 5 + i 16 = i + 1.<br />
(d) 1 2 (1 + i) (1 + i−8 )<br />
Solution: 1 2 (1 + i) (1 + i−8 ) = 1 + i.<br />
1.28 In each case, determine all real x and y which satisfy the given<br />
relation.<br />
19
(a) x + iy = |x − iy|<br />
Proof: Since |x − iy| ≥ 0, we have<br />
(b) x + iy = (x − iy) 2<br />
x ≥ 0 and y = 0.<br />
Proof: Since (x − iy) 2 = x 2 − (2xy) i − y 2 , we have<br />
x = x 2 − y 2 and y = −2xy.<br />
We consider tow cases: (i) y = 0 and (ii) y ≠ 0.<br />
(i) As y = 0 : x = 0 or 1.<br />
(ii) As y ≠ 0 : x = −1/2, and y = ± √ 3<br />
2 .<br />
(c) ∑ 100<br />
k=0 ik = x + iy<br />
Proof: Since ∑ 100<br />
k=0 ik = 1−i101<br />
1−i<br />
= 1−i<br />
1−i<br />
= 1, we have x = 1 and y = 0.<br />
1.29 If z = x+iy, x and y real, the complex conjugate of z is the complex<br />
number ¯z = x − iy. Prove that:<br />
(a) Conjugate of (z 1 + z 2 ) = ¯z 1 + ¯z 2<br />
Proof: Write z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 , then<br />
z 1 + z 2 = (x 1 + x 2 ) + i (y 1 + y 2 )<br />
= (x 1 + x 2 ) − i (y 1 + y 2 )<br />
= (x 1 − iy 1 ) + (x 2 − iy 2 )<br />
= ¯z 1 + ¯z 2 .<br />
(b) z 1 z 2 = ¯z 1¯z 2<br />
Proof: Write z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 , then<br />
z 1 z 2 = (x 1 x 2 − y 1 y 2 ) + i (x 1 y 2 + x 2 y 1 )<br />
= (x 1 x 2 − y 1 y 2 ) − i (x 1 y 2 + x 2 y 1 )<br />
and<br />
¯z 1¯z 2 = (x 1 − iy 1 ) (x 2 − iy 2 )<br />
= (x 1 x 2 − y 1 y 2 ) − i (x 1 y 2 + x 2 y 1 ) .<br />
20
So, z 1 z 2 = ¯z 1¯z 2<br />
(c) z¯z = |z| 2<br />
Proof: Write z = x + iy and thus<br />
z¯z = x 2 + y 2 = |z| 2 .<br />
(d) z + ¯z =twice the real part of z<br />
Proof: Write z = x + iy, then<br />
z + ¯z = 2x,<br />
twice the real part of z.<br />
(e) (z − ¯z) /i =twice the imaginary part of z<br />
Proof: Write z = x + iy, then<br />
z − ¯z<br />
i<br />
= 2y,<br />
twice the imaginary part of z.<br />
1.30 Describe geometrically the set of complex numbers z which satisfies<br />
each of the following conditions:<br />
(a) |z| = 1<br />
Solution: <strong>The</strong> unit circle centered at zero.<br />
(b) |z| < 1<br />
Solution: <strong>The</strong> open unit disk centered at zero.<br />
(c) |z| ≤ 1<br />
Solution: <strong>The</strong> closed unit disk centered at zero.<br />
(d) z + ¯z = 1<br />
Solution: Write z = x + iy, then z + ¯z = 1 means that x = 1/2. So, the<br />
set is the line x = 1/2.<br />
(e) z − ¯z = i<br />
21
Proof: Write z = x + iy, then z − ¯z = i means that y = 1/2. So, the set<br />
is the line y = 1/2.<br />
(f) z + ¯z = |z| 2<br />
Proof: Write z = x + iy, then 2x = x 2 + y 2 ⇔ (x − 1) 2 + y 2 = 1. So, the<br />
set is the unit circle centered at (1, 0) .<br />
1.31 Given three complex numbers z 1 , z 2 , z 3 such that |z 1 | = |z 2 | = |z 3 | =<br />
1 and z 1 + z 2 + z 3 = 0. Show that these numbers are vertices of an equilateral<br />
triangle inscribed in the unit circle with center at the origin.<br />
Proof: It is clear that three numbers are vertices of triangle inscribed in<br />
the unit circle with center at the origin. It remains to show that |z 1 − z 2 | =<br />
|z 2 − z 3 | = |z 3 − z 1 | . In addition, it suffices to show that<br />
Note that<br />
which is equivalent to<br />
which is equivalent to<br />
|z 1 − z 2 | = |z 2 − z 3 | .<br />
|2z 1 + z 3 | = |2z 3 + z 1 | by z 1 + z 2 + z 3 = 0<br />
|2z 1 + z 3 | 2 = |2z 3 + z 1 | 2<br />
(2z 1 + z 3 ) (2¯z 1 + ¯z 3 ) = (2z 3 + z 1 ) (2¯z 3 + ¯z 1 )<br />
which is equivalent to<br />
|z 1 | = |z 3 | .<br />
1.32 If a and b are complex numbers, prove that:<br />
(a) |a − b| 2 ≤ ( 1 + |a| 2) ( 1 + |b| 2)<br />
Proof: Consider<br />
(<br />
1 + |a|<br />
2 ) ( 1 + |b| 2) − |a − b| 2 = (1 + āa) ( 1 + ¯bb ) − (a − b) ( ā − ¯b )<br />
= (1 + āb) ( 1 + a¯b )<br />
= |1 + āb| 2 ≥ 0,<br />
22
so, |a − b| 2 ≤ ( 1 + |a| 2) ( 1 + |b| 2)<br />
(b) If a ≠ 0, then |a + b| = |a| + |b| if, and only if, b/a is real and<br />
nonnegative.<br />
Proof: (⇒)Since |a + b| = |a| + |b| , we have<br />
which implies that<br />
which implies that<br />
which implies that<br />
<strong>The</strong>n<br />
(⇐) Suppose that<br />
|a + b| 2 = (|a| + |b|) 2<br />
Re (āb) = |a| |b| = |ā| |b|<br />
b<br />
a = āb<br />
āa<br />
b<br />
a<br />
āb = |ā| |b|<br />
=<br />
|ā| |b|<br />
|a| 2 ≥ 0.<br />
= k, where k ≥ 0.<br />
|a + b| = |a + ka| = (1 + k) |a| = |a| + k |a| = |a| + |b| .<br />
1.33 If a and b are complex numbers, prove that<br />
|a − b| = |1 − āb|<br />
if, and only if, |a| = 1 or |b| = 1. For which a and b is the inequality<br />
|a − b| < |1 − āb| valid<br />
Proof: (⇔) Since<br />
|a − b| = |1 − āb|<br />
⇔ ( ā − ¯b ) (a − b) = (1 − āb) ( 1 − a¯b )<br />
⇔ |a| 2 + |b| 2 = 1 + |a| 2 |b| 2<br />
⇔ ( |a| 2 − 1 ) ( |b| 2 − 1 ) = 0<br />
⇔ |a| 2 = 1 or |b| 2 = 1.<br />
23
By the preceding, it is easy to know that<br />
|a − b| < |1 − āb| ⇔ 0 < ( |a| 2 − 1 ) ( |b| 2 − 1 ) .<br />
So, |a − b| < |1 − āb| if, and only if, |a| > 1 and |b| > 1. (Or |a| < 1 and<br />
|b| < 1).<br />
1.34 If a and c are real constant, b complex, show that the equation<br />
az¯z + b¯z + ¯bz + c = 0 (a ≠ 0, z = x + iy)<br />
represents a circle in the x − y plane.<br />
Proof: Consider<br />
[ (<br />
z¯z −<br />
b ¯b ¯z −<br />
−a −a z + b ) ] b<br />
=<br />
−a −a<br />
−ac + |b|2<br />
a 2 ,<br />
so, we have<br />
( )∣ b ∣∣∣<br />
2<br />
∣ z − =<br />
−a<br />
−ac + |b|2<br />
a 2 .<br />
Hence, as |b| 2 − ac > 0, it is a circle.<br />
−ac+|b| 2<br />
a 2<br />
< 0, it is not a circle.<br />
Remark: <strong>The</strong> idea is easy from the fact<br />
We square both sides and thus<br />
|z − q| = r.<br />
z¯z − q¯z − ¯qz + ¯qq = r 2 .<br />
As −ac+|b|2<br />
a 2 = 0, it is a point. As<br />
1.35 Recall the definition of the inverse tangent: given a real number t,<br />
tan −1 (t) is the unique real number θ which satisfies the two conditions<br />
− π 2 < θ < +π , tan θ = t.<br />
2<br />
If z = x + iy, show that<br />
(a) arg (z) = tan ( ) −1 y<br />
x , if x > 0<br />
24
Proof: Note that in this text book, we say arg (z) is the principal argument<br />
of z, denoted by θ = arg z, where −π < θ ≤ π.<br />
So, as x > 0, arg z = tan −1 ( y<br />
x)<br />
.<br />
(b) arg (z) = tan −1 ( y<br />
x)<br />
+ π, if x < 0, y ≥ 0<br />
Proof: As x < 0, and y ≥ 0. <strong>The</strong> point (x, y) is lying on S = {(x, y) : x < 0, y ≥ 0} .<br />
Note that −π < arg z ≤ π, so we have arg (z) = tan −1 ( y<br />
x)<br />
+ π.<br />
(c) arg (z) = tan −1 ( y<br />
x)<br />
− π, if x < 0, y < 0<br />
Proof: Similarly for (b). So, we omit it.<br />
(d) arg (z) = π 2 if x = 0, y > 0; arg (z) = − π 2<br />
if x = 0, y < 0.<br />
Proof: It is obvious.<br />
1.36 Define the folowing ”pseudo-ordering” of the complex numbers:<br />
we say z 1 < z 2 if we have either<br />
(i) |z 1 | < |z 2 | or (ii) |z 1 | = |z 2 | and arg (z 1 ) < arg (z 2 ) .<br />
Which of Axioms 6,7,8,9 are satisfied by this relation<br />
Proof: (1) For axiom 6, we prove that it holds as follows. Given z 1 =<br />
r 1 e i arg(z 1) , and r 2 e i arg(z 2) , then if z 1 = z 2 , there is nothing to prove it. If<br />
z 1 ≠ z 2 , there are two possibilities: (a) r 1 ≠ r 2 , or (b) r 1 = r 2 and arg (z 1 ) ≠<br />
arg (z 2 ) . So, it is clear that axiom 6 holds.<br />
(2) For axiom 7, we prove that it does not hold as follows. Given z 1 = 1<br />
and z 2 = −1, then it is clear that z 1 < z 2 since |z 1 | = |z 2 | = 1 and arg (z 1 ) =<br />
0 < arg (z 2 ) = π. However, let z 3 = −i, we have<br />
z 1 + z 3 = 1 − i > z 2 + z 3 = −1 − i<br />
since<br />
and<br />
|z 1 + z 3 | = |z 2 + z 3 | = √ 2<br />
arg (z 1 + z 3 ) = − π 4 > −3π 4 = arg (z 2 + z 3 ) .<br />
(3) For axiom 8, we prove that it holds as follows. If z 1 > 0 and z 2 > 0,<br />
then |z 1 | > 0 and |z 2 | > 0. Hence, z 1 z 2 > 0 by |z 1 z 2 | = |z 1 | |z 2 | > 0.<br />
(4) For axiom 9, we prove that it holds as follows. If z 1 > z 2 and z 2 > z 3 ,<br />
we consider the following cases. Since z 1 > z 2 , we may have (a) |z 1 | > |z 2 | or<br />
(b) |z 1 | = |z 2 | and arg (z 1 ) < arg (z 2 ) .<br />
As |z 1 | > |z 2 | , it is clear that |z 1 | > |z 3 | . So, z 1 > z 3 .<br />
25
As |z 1 | = |z 2 | and arg (z 1 ) < arg (z 2 ) , we have arg (z 1 ) > arg (z 3 ) . So,<br />
z 1 > z 3 .<br />
1.37 Which of Axioms 6,7,8,9 are satisfied if the pseudo-ordering is<br />
defined as follows We say (x 1 , y 1 ) < (x 2 , y 2 ) if we have either (i) x 1 < x 2 or<br />
(ii) x 1 = x 2 and y 1 < y 2 .<br />
Proof: (1) For axiom 6, we prove that it holds as follows. Given x =<br />
(x 1 , y 1 ) and y = (x 2 , y 2 ) . If x = y, there is nothing to prove it. We consider<br />
x ≠ y : As x ≠ y, we have x 1 ≠ x 2 or y 1 ≠ y 2 . Both cases imply x < y or<br />
y < x.<br />
(2) For axiom 7, we prove that it holds as follows. Given x = (x 1 , y 1 ) ,<br />
y = (x 2 , y 2 ) and z = (z 1 , z 3 ) . If x < y, then there are two possibilities: (a)<br />
x 1 < x 2 or (b) x 1 = x 2 and y 1 < y 2 .<br />
For case (a), it is clear that x 1 + z 1 < y 1 + z 1 . So, x + z < y + z.<br />
For case (b), it is clear that x 1 + z 1 = y 1 + z 1 and x 2 + z 2 < y 2 + z 2 . So,<br />
x + z < y + z.<br />
(3) For axiom 8, we prove that it does not hold as follows. Consider<br />
x = (1, 0) and y = (0, 1) , then it is clear that x > 0 and y > 0. However,<br />
xy = (0, 0) = 0.<br />
(4) For axiom 9, we prove that it holds as follows. Given x = (x 1 , y 1 ) ,<br />
y = (x 2 , y 2 ) and z = (z 1 , z 3 ) . If x > y and y > z, then we consider the<br />
following cases. (a) x 1 > y 1 , or (b) x 1 = y 1 .<br />
For case (a), it is clear that x 1 > z 1 . So, x > z.<br />
For case (b), it is clear that x 2 > y 2 . So, x > z.<br />
1.38 State and prove a theorem analogous to <strong>The</strong>orem 1.48, expressing<br />
arg (z 1 /z 2 ) in terms of arg (z 1 ) and arg (z 2 ) .<br />
Proof: Write z 1 = r 1 e i arg(z 1) and z 2 = r 2 e i arg(z 2) , then<br />
z 1<br />
z 2<br />
= r 1<br />
r 2<br />
e i[arg(z 1)−arg(z 2 )] .<br />
Hence,<br />
where<br />
( )<br />
z1<br />
arg = arg (z 1 ) − arg (z 2 ) + 2πn (z 1 , z 2 ) ,<br />
z 2<br />
⎧<br />
⎨ 0 if − π < arg (z 1 ) − arg (z 2 ) ≤ π<br />
n (z 1 , z 2 ) = 1 if − 2π < arg (z 1 ) − arg (z 2 ) ≤ −π<br />
⎩<br />
−1 if π < arg (z 1 ) − arg (z 2 ) < 2π<br />
.<br />
26
1.39 State and prove a theorem analogous to <strong>The</strong>orem 1.54, expressing<br />
Log (z 1 /z 2 ) in terms of Log (z 1 ) and Log (z 2 ) .<br />
Proof: Write z 1 = r 1 e i arg(z 1) and z 2 = r 2 e i arg(z 2) , then<br />
Hence,<br />
∣ Log (z 1 /z 2 ) = log<br />
z 1 ∣∣∣<br />
∣ + i arg<br />
z 2<br />
z 1<br />
z 2<br />
= r 1<br />
r 2<br />
e i[arg(z 1)−arg(z 2 )] .<br />
(<br />
z1<br />
z 2<br />
)<br />
= log |z 1 | − log |z 2 | + i [arg (z 1 ) − arg (z 2 ) + 2πn (z 1 , z 2 )] by xercise 1.38<br />
= Log (z 1 ) − Log (z 2 ) + i2πn (z 1 , z 2 ) .<br />
1.40 Prove that the nth roots of 1 (also called the nth roots of unity)<br />
are given by α, α 2 , ..., α n , where α = e 2πi/n , and show that the roots ≠ 1<br />
satisfy the equation<br />
1 + x + x 2 + ... + x n−1 = 0.<br />
Proof: By <strong>The</strong>orem 1.51, we know that the roots of 1 are given by<br />
α, α 2 , ..., α n , where α = e 2πi/n . In addition, since<br />
which implies that<br />
x n = 1 ⇒ (x − 1) ( 1 + x + x 2 + ... + x n−1) = 0<br />
1 + x + x 2 + ... + x n−1 = 0 if x ≠ 1.<br />
So, all roots except 1 satisfy the equation<br />
1 + x + x 2 + ... + x n−1 = 0.<br />
1.41 (a) Prove that |z i | < e π for all complex z ≠ 0.<br />
Proof: Since<br />
z i = e iLog(z) = e − arg(z)+i log|z| ,<br />
27
we have<br />
∣ ∣z i ∣ ∣ = e<br />
− arg(z) < e π<br />
by −π < arg (z) ≤ π.<br />
(b) Prove that there is no constant M > 0 such that |cos z| < M for all<br />
complex z.<br />
Proof: Write z = x + iy and thus,<br />
cos z = cos x cosh y − i sin x sinh y<br />
which implies that<br />
|cos x cosh y| ≤ |cos z| .<br />
Let x = 0 and y be real, then<br />
e y<br />
2 ≤ 1 ∣ e y + e −y∣ ∣ ≤ |cos z| .<br />
2<br />
So, there is no constant M > 0 such that |cos z| < M for all complex z.<br />
Remark: <strong>The</strong>re is an important theorem related with this exercise. We<br />
state it as a reference. (Liouville’s <strong>The</strong>orem) A bounded entire function<br />
is constant. <strong>The</strong> reader can see the book, <strong>Complex</strong> Analysis by Joseph<br />
Bak, and Donald J. Newman, pp 62-63. Liouville’s <strong>The</strong>orem can<br />
be used to prove the much important theorem, Fundamental <strong>The</strong>orem of<br />
Algebra.<br />
1.42 If w = u + iv (u, v real), show that<br />
z w = e u log|z|−v arg(z) e i[v log|z|+u arg(z)] .<br />
Proof: Write z w = e wLog(z) , and thus<br />
wLog (z) = (u + iv) (log |z| + i arg (z))<br />
= [u log |z| − v arg (z)] + i [v log |z| + u arg (z)] .<br />
So,<br />
z w = e u log|z|−v arg(z) e i[v log|z|+u arg(z)] .<br />
28
1.43 (a) Prove that Log (z w ) = wLog z +2πin.<br />
Proof: Write w = u + iv, where u and v are real. <strong>The</strong>n<br />
Log (z w ) = log |z w | + i arg (z w )<br />
= log [ e u log|z|−v arg(z)] + i [v log |z| + u arg (z)] + 2πin by Exercise1.42<br />
= u log |z| − v arg (z) + i [v log |z| + u arg (z)] + 2πin.<br />
On the other hand,<br />
wLogz + 2πin = (u + iv) (log |z| + i arg (z)) + 2πin<br />
= u log |z| − v arg (z) + i [v log |z| + u arg (z)] + 2πin.<br />
Hence, Log (z w ) = wLog z +2πin.<br />
Remark: <strong>The</strong>re is another proof by considering<br />
e Log(zw) = z w = e wLog(z)<br />
which implies that<br />
Log (z w ) = wLogz + 2πin<br />
for some n ∈ Z.<br />
(b) Prove that (z w ) α = z wα e 2πinα , where n is an integer.<br />
Proof: By (a), we have<br />
(z w ) α = e αLog(zw) = e α(wLogz+2πin) = e αwLogz e 2πinα = z αw e 2πinα ,<br />
where n is an integer.<br />
1.44 (i) If θ and a are real numbers, −π < θ ≤ π, prove that<br />
(cos θ + i sin θ) a = cos (aθ) + i sin (aθ) .<br />
Proof: Write cos θ + i sin θ = z, we then have<br />
(cos θ + i sin θ) a = z a = e aLogz = e a[log|eiθ |+i arg(e iθ )] = e<br />
iaθ<br />
= cos (aθ) + i sin (aθ) .<br />
29
Remark: Compare with the Exercise 1.43-(b).<br />
(ii) Show that, in general, the restriction −π < θ ≤ π is necessary in (i)<br />
by taking θ = −π, a = 1 2 .<br />
Proof: As θ = −π, and a = 1 , we have<br />
2<br />
(<br />
(−1) 1 1<br />
2 = e 2 Log(−1) = e π −π<br />
2 i = i ≠ −i = cos<br />
2<br />
) ( −π<br />
+ i sin<br />
2<br />
)<br />
.<br />
(iii) If a is an integer, show that the formula in (i) holds without any<br />
restriction on θ. In this case it is known as DeMorvre’s theorem.<br />
Proof: By Exercise 1.43, as a is an integer we have<br />
(z w ) a = z wa ,<br />
where z w = e iθ . <strong>The</strong>n<br />
(<br />
e<br />
iθ ) a<br />
= e iθa = cos (aθ) + i sin (aθ) .<br />
1.45 Use DeMorvre’s theorem (Exercise 1.44) to derive the triginometric<br />
identities<br />
sin 3θ = 3 cos 2 θ sin θ − sin 3 θ<br />
cos 3θ = cos 3 θ − 3 cos θ sin 2 θ,<br />
valid for real θ. Are these valid when θ is complex<br />
Proof: By Exercise 1.44-(iii), we have for any real θ,<br />
By Binomial <strong>The</strong>orem, we have<br />
(cos θ + i sin θ) 3 = cos (3θ) + i sin (3θ) .<br />
sin 3θ = 3 cos 2 θ sin θ − sin 3 θ<br />
and<br />
cos 3θ = cos 3 θ − 3 cos θ sin 2 θ.<br />
30
For complex θ, we show that it holds as follows. Note that sin z = eiz −e −iz<br />
2i<br />
and cos z = eiz +e −iz<br />
, we have<br />
2<br />
( ) e<br />
3 cos 2 z sin z − sin 3 iz + e −iz 2 ( ) ( ) e iz − e −iz e iz − e −iz 3<br />
z = 3<br />
−<br />
2<br />
2i<br />
2i<br />
( ) ( ) e 2zi + e −2zi + 2 e iz − e −iz<br />
= 3<br />
+ e3zi − 3e iz + 3e −iz − e −3zi<br />
4<br />
2i<br />
8i<br />
= 1 [ (<br />
3 e 2zi + e −2zi + 2 ) ( e zi − e −zi) + ( e 3zi − 3e iz + 3e −iz − e −3zi)]<br />
8i<br />
= 1 [(<br />
3e 3zi + 3e iz − 3e −iz − 3e −3zi) + ( e 3zi − 3e iz + 3e −iz − e −3zi)]<br />
8i<br />
= 4 (<br />
e 3zi − e −3zi)<br />
8i<br />
= 1 (<br />
e 3zi − e −3zi)<br />
2i<br />
Similarly, we also have<br />
= sin 3z.<br />
cos 3 z − 3 cos z sin 2 z = cos 3z.<br />
1.46 Define tan z = sin z/ cos z and show that for z = x + iy, we have<br />
tan z =<br />
sin 2x + i sinh 2y<br />
cos 2x + cosh 2y .<br />
31
Proof: Since<br />
tan z = sin z sin (x + iy) sin x cosh y + i cos x sinh y<br />
= =<br />
cos z cos (x + iy) cos x cosh y − i sin x sinh y<br />
(sin x cosh y + i cos x sinh y) (cos x cosh y + i sin x sinh y)<br />
=<br />
(cos x cosh y − i sin x sinh y) (cos x cosh y + i sin x sinh y)<br />
(<br />
sin x cos x cosh 2 y − sin x cos x sinh 2 y ) + i ( sin 2 x cosh y sinh y + cos 2 x cosh y sinh y )<br />
=<br />
(cos x cosh y) 2 − (i sin x sinh y) 2<br />
= sin x cos x ( cosh 2 y − sinh 2 y ) + i (cosh y sinh y)<br />
cos 2 x cosh 2 y + sin 2 x sinh 2 since sin 2 x + cos 2 x = 1<br />
y<br />
(sin x cos x) + i (cosh y sinh y)<br />
=<br />
cos 2 x + sinh 2 since cosh 2 y = 1 + sinh 2 y<br />
y<br />
1<br />
2<br />
=<br />
sin 2x + i sinh 2y<br />
2<br />
cos 2 x + sinh 2 since 2 cosh y sinh y = sinh 2y and 2 sin x cos x = sin 2x<br />
y<br />
sin 2x + i sinh 2y<br />
=<br />
2 cos 2 x + 2 sinh 2 y<br />
sin 2x + i sinh 2y<br />
=<br />
2 cos 2 x − 1 + 2 sinh 2 y + 1<br />
sin 2x + i sinh 2y<br />
=<br />
cos 2x + cosh 2y since cos 2x = 2 cos2 x − 1 and 2 sinh 2 y + 1 = cosh 2y.<br />
1.47 Let w be a given complex number. If w ≠ ±1, show that there exists<br />
two values of z = x + iy satisfying the conditions cos z = w and −π < x ≤ π.<br />
Find these values when w = i and when w = 2.<br />
Proof: Since cos z = eiz +e −iz<br />
, if we let e iz = u, then cos z = w implies<br />
2<br />
that<br />
w = u2 + 1<br />
⇒ u 2 − 2wu + 1 = 0<br />
2u<br />
which implies that<br />
So, by <strong>The</strong>orem 1.51,<br />
(u − w) 2 = w 2 − 1 ≠ 0 since w ≠ ±1.<br />
e iz = u = w + ∣ w 2 − 1 ∣ 1/2 e iφ k<br />
, where φ k = arg (w2 − 1)<br />
2<br />
(<br />
)<br />
= w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1)<br />
2<br />
+ 2πk , k = 0, 1.<br />
2<br />
32
So,<br />
ix−y = i (x + iy) = iz = log<br />
∣ w ± ∣ w 2 − 1 ∣ ⎛<br />
(<br />
)⎞<br />
1/2 e i arg (w 2 −1)<br />
2<br />
∣ +i arg ⎝w ± ∣ w 2 − 1 ∣ 1/2 e i arg ( w 2 −1 )<br />
2<br />
⎠<br />
Hence, there exists two values of z = x+iy satisfying the conditions cos z = w<br />
and<br />
⎛<br />
(<br />
)⎞<br />
−π < x = arg ⎝w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1)<br />
2<br />
⎠ ≤ π.<br />
For w = i, we have<br />
(<br />
iz = log ∣ 1 ± √ )<br />
2 i∣ + i arg<br />
which implies that<br />
z = arg<br />
((<br />
1 ± √ ) )<br />
2 i<br />
For w = 2, we have<br />
∣<br />
iz = log ∣2 ± √ 3∣ + i arg<br />
((<br />
1 ± √ ) )<br />
2 i<br />
(<br />
− i log ∣ 1 ± √ )<br />
2 i∣ .<br />
(<br />
2 ± √ )<br />
3<br />
which implies that<br />
(<br />
z = arg 2 ± √ )<br />
3<br />
∣<br />
− i log ∣2 ± √ 3∣ .<br />
1.48 Prove Lagrange’s identity for complex numbers:<br />
∣ n∑ ∣∣∣∣<br />
2 n∑ n∑<br />
a k b k = |a k | 2 |b k | 2 − ∑ ( ) 2<br />
ak¯bj − ā j b k .<br />
∣<br />
k=1<br />
k=1<br />
k=1<br />
1≤k
Proof: By Exercise 1.44 (i), we have<br />
sin nθ =<br />
[ n+1<br />
2 ]<br />
∑<br />
k=1<br />
( n2k−1<br />
)<br />
sin 2k−1 θ cos n−(2k−1) θ<br />
⎧<br />
⎫<br />
⎪⎨ [ n+1<br />
2<br />
∑<br />
]<br />
(<br />
= sin n θ<br />
n2k−1<br />
)<br />
⎪⎬<br />
cot n−(2k−1) θ<br />
⎪⎩<br />
⎪⎭<br />
k=1<br />
= sin n θ { ( n 1) cot n−1 θ − ( n 3) cot n−3 θ + ( n 5) cot n−5 θ − +... } .<br />
(b) If 0 < θ < π/2, prove that<br />
sin (2m + 1) θ = sin 2m+1 θP m<br />
(<br />
cot 2 θ )<br />
where P m is the polynomial of degree m given by<br />
P m (x) = ( )<br />
2m+1<br />
1 x m − ( )<br />
2m+1<br />
3 x m−1 + ( )<br />
2m+1<br />
5 x m−2 − +...<br />
Use this to show that P m has zeros at the m distinct points x k = cot 2 {πk/ (2m + 1)}<br />
for k = 1, 2, ..., m.<br />
Proof: By (a),<br />
sin (2m + 1) θ<br />
= sin 2m+1 θ<br />
{ (2m+1<br />
1<br />
) (<br />
cot 2 θ ) m<br />
−<br />
( 2m+1<br />
3<br />
= sin 2m+1 θP m<br />
(<br />
cot 2 θ ) , where P m (x) =<br />
) (<br />
cot 2 θ ) m−1 (<br />
+<br />
2m+1<br />
) (<br />
5 cot 2 θ ) }<br />
m−2<br />
− +...<br />
m+1<br />
∑<br />
k=1<br />
( 2m+1<br />
)<br />
2k−1 x m+1−k . (*)<br />
In addition, by (*), sin (2m + 1) θ = 0 if, and only if, P m (cot 2 θ) = 0. Hence,<br />
P m has zeros at the m distinct points x k = cot 2 {πk/ (2m + 1)} for k =<br />
1, 2, ..., m.<br />
(c) Show that the sum of the zeros of P m is given by<br />
m∑<br />
cot 2 πk<br />
2m + 1<br />
k=1<br />
34<br />
m (2m − 1)<br />
= ,<br />
3
and the sum of their squares is given by<br />
m∑<br />
cot 4 πk<br />
2m + 1 = m (2m − 1) (4m2 + 10m − 9)<br />
.<br />
45<br />
k=1<br />
Note. <strong>The</strong>re identities can be used to prove that ∑ ∞<br />
∑ n=1 n−2 = π 2 /6 and<br />
∞<br />
n=1 n−4 = π 4 /90. (See Exercises 8.46 and 8.47.)<br />
Proof: By (b), we know that sum of the zeros of P m is given by<br />
( ( m∑ m∑<br />
x k = cot 2 πk − 2m+1<br />
))<br />
2m + 1 = − ( 3 m (2m − 1)<br />
2m+1<br />
) = .<br />
3<br />
k=1<br />
k=1<br />
<strong>And</strong> the sum of their squares is given by<br />
m∑ m∑<br />
x 2 k = cot 4 πk<br />
2m + 1<br />
k=1 k=1<br />
( m<br />
) 2 (<br />
∑<br />
∑<br />
= x k − 2<br />
k=1<br />
1<br />
1≤i
So, let z = 1, we obtain that<br />
n−1<br />
∏ (<br />
n =<br />
) n−1<br />
∏<br />
[(<br />
1 − e<br />
2πik/n<br />
= 1 − cos 2πk<br />
n<br />
k=1<br />
n−1<br />
∏<br />
=<br />
k=1<br />
n−1<br />
∏<br />
=<br />
k=1<br />
n−1<br />
∏<br />
= 2 n−1<br />
(<br />
2 sin 2 πk<br />
n<br />
(<br />
2 sin πk<br />
n<br />
k=1<br />
n−1<br />
∏<br />
= 2 n−1<br />
=<br />
[<br />
k=1<br />
∏<br />
2 n−1 n−1<br />
k=1<br />
n−1<br />
∏<br />
= 2 n−1<br />
k=1<br />
)<br />
− i<br />
(<br />
sin πk<br />
n<br />
(<br />
sin πk<br />
n<br />
k=1<br />
) (<br />
sin πk<br />
n<br />
(<br />
2 sin πk πk<br />
cos<br />
n n<br />
)<br />
) (<br />
cos<br />
)<br />
e i( 3π 2 + πk<br />
n )<br />
(<br />
sin πk ) ] e ∑ n−1 3π<br />
k=1 2 + πk<br />
n<br />
n<br />
(<br />
sin πk<br />
n<br />
)<br />
.<br />
)<br />
)<br />
πk<br />
− i cos<br />
n<br />
( 3π<br />
2 + πk )<br />
+ i sin<br />
n<br />
(<br />
− i sin 2πk )]<br />
n<br />
( 3π<br />
2 + πk ))<br />
n<br />
36
Some Basic Notations Of Set <strong>The</strong>ory<br />
References<br />
<strong>The</strong>re are some good books about set theory; we write them down. We<br />
wish the reader can get more.<br />
1. Set <strong>The</strong>ory and Related Topics by Seymour Lipschutz.<br />
2. Set <strong>The</strong>ory by Charles C. Pinter.<br />
3. <strong>The</strong>ory of sets by Kamke.<br />
4. Naive set by Halmos.<br />
2.1 Prove <strong>The</strong>orem 2.2. Hint. (a, b) = (c, d) means {{a} , {a, b}} =<br />
{{c} , {c, d}} . Now appeal to the definition of set equality.<br />
Proof: (⇐) It is trivial.<br />
(⇒) Suppose that (a, b) = (c, d) , it means that {{a} , {a, b}} = {{c} , {c, d}} .<br />
It implies that<br />
{a} ∈ {{c} , {c, d}} and {a, b} ∈ {{c} , {c, d}} .<br />
So, if a ≠ c, then {a} = {c, d} . It implies that c ∈ {a} which is impossible.<br />
Hence, a = c. Similarly, we have b = d.<br />
2.2 Let S be a relation and let D (S) be its domain. <strong>The</strong> relation S is<br />
said to be<br />
(i) reflexive if a ∈ D (S) implies (a, a) ∈ S,<br />
(ii) symmetric if (a, b) ∈ S implies (b, a) ∈ S,<br />
(iii) transitive if (a, b) ∈ S and (b, c) ∈ S implies (a, c) ∈ S.<br />
A relation which is symmetric, reflexive, and transitive is called an equivalence<br />
relation. Determine which of these properties is possessed by S, if S<br />
is the set of all pairs of real numbers (x, y) such that<br />
(a) x ≤ y<br />
Proof: Write S = {(x, y) : x ≤ y} , then we check that (i) reflexive, (ii)<br />
symmetric, and (iii) transitive as follows. It is clear that D (S) = R.<br />
1
(i) Since x ≤ x, (x, x) ∈ S. That is, S is reflexive.<br />
(ii) If (x, y) ∈ S, i.e., x ≤ y, then y ≤ x. So, (y, x) ∈ S. That is, S is<br />
symmetric.<br />
(iii) If (x, y) ∈ S and (y, z) ∈ S, i.e., x ≤ y and y ≤ z, then x ≤ z. So,<br />
(x, z) ∈ S. That is, S is transitive.<br />
(b) x < y<br />
Proof: Write S = {(x, y) : x < y} , then we check that (i) reflexive, (ii)<br />
symmetric, and (iii) transitive as follows. It is clear that D (S) = R.<br />
(i) It is clear that for any real x, we cannot have x < x. So, S is not<br />
reflexive.<br />
(ii) It is clear that for any real x, and y, we cannot have x < y and y < x<br />
at the same time. So, S is not symmetric.<br />
(iii) If (x, y) ∈ S and (y, z) ∈ S, then x < y and y < z. So, x < z wich<br />
implies (x, z) ∈ S. That is, S is transitive.<br />
(c) x < |y|<br />
Proof: Write S = {(x, y) : x < |y|} , then we check that (i) reflexive, (ii)<br />
symmetric, and (iii) transitive as follows. It is clear that D (S) = R.<br />
(i) Since it is impossible for 0 < |0| , S is not reflexive.<br />
(ii) Since (−1, 2) ∈ S but (2, −1) /∈ S, S is not symmetric.<br />
(iii) Since (0, −1) ∈ S and (−1, 0) ∈ S, but (0, 0) /∈ S, S is not transitive.<br />
(d) x 2 + y 2 = 1<br />
Proof: Write S = {(x, y) : x 2 + y 2 = 1} , then we check that (i) reflexive,<br />
(ii) symmetric, and (iii) transitive as follows. It is clear that D (S) = [−1, 1] ,<br />
an closed interval with endpoints, −1 and 1.<br />
(i) Since 1 ∈ D (S) , and it is impossible for (1, 1) ∈ S by 1 2 + 1 2 ≠ 1, S<br />
is not reflexive.<br />
(ii) If (x, y) ∈ S, then x 2 + y 2 = 1. So, (y, x) ∈ S. That is, S is symmetric.<br />
(iii) Since (1, 0) ∈ S and (0, 1) ∈ S, but (1, 1) /∈ S, S is not transitive.<br />
(e) x 2 + y 2 < 0<br />
Proof: Write S = {(x, y) : x 2 + y 2 < 1} = φ, then S automatically satisfies<br />
(i) reflexive, (ii) symmetric, and (iii) transitive.<br />
(f) x 2 + x = y 2 + y<br />
Proof: Write S = {(x, y) : x 2 + x = y 2 + y} = {(x, y) : (x − y) (x + y − 1) = 0} ,<br />
then we check that (i) reflexive, (ii) symmetric, and (iii) transitive as follows.<br />
It is clear that D (S) = R.<br />
2
(i) If x ∈ R, it is clear that (x, x) ∈ S. So, S is reflexive.<br />
(ii) If (x, y) ∈ S, it is clear that (y, x) ∈ S. So, S is symmetric.<br />
(iii) If (x, y) ∈ S and (y, z) ∈ S, it is clear that (x, z) ∈ S. So, S is<br />
transitive.<br />
2.3 <strong>The</strong> following functions F and G are defined for all real x by the<br />
equations given. In each case where the composite function G ◦ F can be<br />
formed, give the domain of G◦F and a formula (or formulas) for (G ◦ F ) (x) .<br />
(a) F (x) = 1 − x, G (x) = x 2 + 2x<br />
Proof: Write<br />
G ◦ F (x) = G [F (x)] = G [1 − x] = (1 − x) 2 + 2 (1 − x) = x 2 − 4x + 3.<br />
It is clear that the domain of G ◦ F (x) is R.<br />
(b) F (x) = x + 5, G (x) = |x| /x if x ≠ 0, G (0) = 0.<br />
Proof: Write<br />
G ◦ F (x) = G [F (x)] =<br />
{<br />
G (x + 5) =<br />
|x+5|<br />
if x ≠ −5.<br />
x+5<br />
0 if x = −5.<br />
It is clear that the domain of G ◦ F (x) is R.<br />
{ { 2x, if 0 ≤ x ≤ 1<br />
x<br />
(c) F (x) =<br />
G (x) =<br />
2 , if 0 ≤ x ≤ 1<br />
1, otherwise,<br />
0, otherwise.<br />
Proof: Write<br />
⎧<br />
⎨ 4x 2 if x ∈ [0, 1/2]<br />
G ◦ F (x) = G [F (x)] = 0 if x ∈ (1/2, 1]<br />
⎩<br />
1 if x ∈ R − [0, 1]<br />
It is clear that the domain of G ◦ F (x) is R.<br />
Find F (x) if G (x) and G [F (x)] are given as follows:<br />
(d) G (x) = x 3 , G [F (x)] = x 3 − 3x 2 + 3x − 1.<br />
Proof: With help of (x − 1) 3 = x 3 − 3x 2 + 3x − 1, it is easy to know<br />
that F (x) = 1 − x. In addition, there is not other function H (x) such that<br />
G [H (x)] = x 3 − 3x 2 + 3x − 1 since G (x) = x 3 is 1-1.<br />
(e) G (x) = 3 + x + x 2 , G [F (x)] = x 2 − 3x + 5.<br />
3<br />
.
Proof: Write G (x) = ( x + 1 2) 2<br />
+<br />
11<br />
4 , then<br />
G [F (x)] =<br />
(<br />
F (x) + 1 ) 2<br />
+ 11<br />
2 4 = x2 − 3x + 5<br />
which implies that<br />
which implies that<br />
(2F (x) + 1) 2 = (2x − 3) 2<br />
F (x) = x − 2 or − x + 1.<br />
2.4 Given three functions F, G, H, what restrictions must be placed on<br />
their domains so that the following four composite functions can be defined<br />
G ◦ F, H ◦ G, H ◦ (G ◦ F ) , (H ◦ G) ◦ F.<br />
Proof: It is clear for answers,<br />
R (F ) ⊆ D (G) and R (G) ⊆ D (H) .<br />
Assuming that H ◦ (G ◦ F ) and (H ◦ G) ◦ F can be defined, prove that<br />
associative law:<br />
H ◦ (G ◦ F ) = (H ◦ G) ◦ F.<br />
Proof: Given any x ∈ D (F ) , then<br />
((H ◦ G) ◦ F ) (x) = (H ◦ G) (F (x))<br />
= H (G (F (x)))<br />
= H ((G ◦ F ) (x))<br />
= (H ◦ (G ◦ F )) (x) .<br />
So, H ◦ (G ◦ F ) = (H ◦ G) ◦ F.<br />
2.5 Prove the following set-theoretic identities for union and intersection:<br />
4
(a) A ∪ (B ∪ C) = (A ∪ B) ∪ C, A ∩ (B ∩ C) = (A ∩ B) ∩ C.<br />
Proof: For the part, A∪(B ∪ C) = (A ∪ B)∪C : Given x ∈ A∪(B ∪ C) ,<br />
we have x ∈ A or x ∈ B ∪ C. That is, x ∈ A or x ∈ B or x ∈ C. Hence,<br />
x ∈ A∪B or x ∈ C. It implies x ∈ (A ∪ B)∪C. Similarly, if x ∈ (A ∪ B)∪C,<br />
then x ∈ A ∪ (B ∪ C) . <strong>The</strong>refore, A ∪ (B ∪ C) = (A ∪ B) ∪ C.<br />
For the part, A∩(B ∩ C) = (A ∩ B)∩C : Given x ∈ A∩(B ∩ C) , we have<br />
x ∈ A and x ∈ B ∩C. That is, x ∈ A and x ∈ B and x ∈ C. Hence, x ∈ A∩B<br />
and x ∈ C. It implies x ∈ (A ∩ B) ∩ C. Similarly, if x ∈ (A ∩ B) ∩ C, then<br />
x ∈ A ∩ (B ∩ C) . <strong>The</strong>refore, A ∩ (B ∩ C) = (A ∩ B) ∩ C.<br />
(b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) .<br />
Proof: Given x ∈ A ∩ (B ∪ C) , then x ∈ A and x ∈ B ∪ C. We consider<br />
two cases as follows.<br />
If x ∈ B, then x ∈ A ∩ B. So, x ∈ (A ∩ B) ∪ (A ∩ C) .<br />
If x ∈ C, then x ∈ A ∩ C. So, x ∈ (A ∩ B) ∪ (A ∩ C) .<br />
So, we have shown that<br />
A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) . (*)<br />
Conversely, given x ∈ (A ∩ B) ∪ (A ∩ C) , then x ∈ A ∩ B or x ∈ A ∩ C.<br />
We consider two cases as follows.<br />
If x ∈ A ∩ B, then x ∈ A ∩ (B ∪ C) .<br />
If x ∈ A ∩ C, then x ∈ A ∩ (B ∪ C) .<br />
So, we have shown that<br />
(A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) . (**)<br />
By (*) and (**), we have proved it.<br />
(c) (A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C)<br />
Proof: Given x ∈ (A ∪ B) ∩ (A ∪ C) , then x ∈ A ∪ B and x ∈ A ∪ C.<br />
We consider two cases as follows.<br />
If x ∈ A, then x ∈ A ∪ (B ∩ C) .<br />
If x /∈ A, then x ∈ B and x ∈ C. So, x ∈ B ∩ C. It implies that<br />
x ∈ A ∪ (B ∩ C) .<br />
<strong>The</strong>refore, we have shown that<br />
(A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C) . (*)<br />
5
Conversely, if x ∈ A ∪ (B ∩ C) , then x ∈ A or x ∈ B ∩ C. We consider<br />
two cases as follows.<br />
If x ∈ A, then x ∈ (A ∪ B) ∩ (A ∪ C) .<br />
If x ∈ B ∩ C, then x ∈ A ∪ B and x ∈ A ∪ C. So, x ∈ (A ∪ B) ∩ (A ∪ C) .<br />
<strong>The</strong>refore, we have shown that<br />
A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C) . (*)<br />
By (*) and (**), we have proved it.<br />
(d) (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) = (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C)<br />
Proof: Given x ∈ (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) , then<br />
x ∈ A ∪ B and x ∈ B ∪ C and x ∈ C ∪ A. (*)<br />
We consider the cases to show x ∈ (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) as follows.<br />
For the case (x ∈ A):<br />
If x ∈ B, then x ∈ A ∩ B.<br />
If x /∈ B, then by (*), x ∈ C. So, x ∈ A ∩ C.<br />
Hence, in this case, we have proved that x ∈ (A ∩ B)∪(A ∩ C)∪(B ∩ C) .<br />
For the case (x /∈ A):<br />
If x ∈ B, then by (*), x ∈ C. So, x ∈ B ∩ C.<br />
If x /∈ B, then by (*), it is impossible.<br />
Hence, in this case, we have proved that x ∈ (A ∩ B)∪(A ∩ C)∪(B ∩ C) .<br />
From above,<br />
(A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) ⊆ (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C)<br />
Similarly, we also have<br />
(A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) .<br />
So, we have proved it.<br />
Remark: <strong>The</strong>re is another proof, we write it as a reference.<br />
Proof: Consider<br />
(A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A)<br />
= [(A ∪ B) ∩ (B ∪ C)] ∩ (C ∪ A)<br />
= [B ∪ (A ∩ C)] ∩ (C ∪ A)<br />
= [B ∩ (C ∪ A)] ∪ [(A ∩ C) ∩ (C ∪ A)]<br />
= [(B ∩ C) ∪ (B ∩ A)] ∪ (A ∩ C)<br />
= (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C) .<br />
6
(e) A ∩ (B − C) = (A ∩ B) − (A ∩ C)<br />
Proof: Given x ∈ A ∩ (B − C) , then x ∈ A and x ∈ B − C. So, x ∈ A<br />
and x ∈ B and x /∈ C. So, x ∈ A ∩ B and x /∈ C. Hence,<br />
x ∈ (A ∩ B) − C ⊆ (A ∩ B) − (A ∩ C) . (*)<br />
Conversely, given x ∈ (A ∩ B) − (A ∩ C) , then x ∈ A ∩ B and x /∈ A ∩ C.<br />
So, x ∈ A and x ∈ B and x /∈ C. So, x ∈ A and x ∈ B − C. Hence,<br />
By (*) and (**), we have proved it.<br />
(f) (A − C) ∩ (B − C) = (A ∩ B) − C<br />
x ∈ A ∩ (B − C) (**)<br />
Proof: Given x ∈ (A − C) ∩ (B − C) , then x ∈ A − C and x ∈ B − C.<br />
So, x ∈ A and x ∈ B and x /∈ C. So, x ∈ (A ∩ B) − C. Hence,<br />
(A − C) ∩ (B − C) ⊆ (A ∩ B) − C. (*)<br />
Conversely, given x ∈ (A ∩ B)−C, then x ∈ A and x ∈ B and x /∈ C. Hence,<br />
x ∈ A − C and x ∈ B − C. Hence,<br />
By (*) and (**), we have proved it.<br />
(A ∩ B) − C ⊆ (A − C) ∩ (B − C) . (**)<br />
(g) (A − B) ∪ B = A if, and only if, B ⊆ A<br />
Proof: (⇒) Suppose that (A − B) ∪ B = A, then it is clear that B ⊆ A.<br />
(⇐) Suppose that B ⊆ A, then given x ∈ A, we consider two cases.<br />
If x ∈ B, then x ∈ (A − B) ∪ B.<br />
If x /∈ B, then x ∈ A − B. Hence, x ∈ (A − B) ∪ B.<br />
From above, we have<br />
A ⊆ (A − B) ∪ B.<br />
In addition, it is obviously (A − B) ∪ B ⊆ A since A − B ⊆ A and B ⊆ A.<br />
2.6 Let f : S → T be a function. If A and B are arbitrary subsets of S,<br />
prove that<br />
f (A ∪ B) = f (A) ∪ f (B) and f (A ∩ B) ⊆ f (A) ∩ f (B) .<br />
7
Generalize to arbitrary unions and intersections.<br />
Proof: First, we prove f (A ∪ B) = f (A) ∪ f (B) as follows. Let y ∈<br />
f (A ∪ B) , then y = f (a) or y = f (b) , where a ∈ A and b ∈ B. Hence,<br />
y ∈ f (A) ∪ f (B) . That is,<br />
f (A ∪ B) ⊆ f (A) ∪ f (B) .<br />
Conversely, if y ∈ f (A) ∪ f (B) , then y = f (a) or y = f (b) , where a ∈ A<br />
and b ∈ B. Hence, y ∈ f (A ∪ B) . That is,<br />
f (A) ∪ f (B) ⊆ f (A ∪ B) .<br />
So, we have proved that f (A ∪ B) = f (A) ∪ f (B) .<br />
For the part f (A ∩ B) ⊆ f (A) ∩ f (B) : Let y ∈ f (A ∩ B) , then y =<br />
f (x) , where x ∈ A∩B. Hence, y ∈ f (A) and y ∈ f (B) . That is, f (A ∩ B) ⊆<br />
f (A) ∩ f (B) .<br />
For arbitrary unions and intersections, we have the following facts, and<br />
the proof is easy from above. So, we omit the detail.<br />
<strong>And</strong><br />
f (∪ i∈I A i ) = ∪ i∈I f (A i ) , where I is an index set.<br />
f (∩ i∈I A i ) ⊆ ∩ i∈I f (A i ) , where I is an index set.<br />
Remark: We should note why the equality does NOT hold for the case<br />
of intersection. for example, consider A = {1, 2} and B = {1, 3} , where<br />
f (1) = 1 and f (2) = 2 and f (3) = 2.<br />
f (A ∩ B) = f ({1}) = {1} ⊆ {1, 2} ⊆ f ({1, 2}) ∩ f ({1, 3}) = f (A) ∩ f (B) .<br />
2.7 Let f : S → T be a function. If Y ⊆ T, we denote by f −1 (Y ) the<br />
largest subset of S which f maps into Y. That is,<br />
f −1 (Y ) = {x : x ∈ S and f (x) ∈ Y } .<br />
<strong>The</strong> set f −1 (Y ) is called the inverse image of Y under f. Prove that the<br />
following for arbitrary subsets X of S and Y of T.<br />
(a) X ⊆ f −1 [f (X)]<br />
8
Proof: Given x ∈ X, then f (x) ∈ f (X) . Hence, x ∈ f −1 [f (X)] by<br />
definition of the inverse image of f (X) under f. So, X ⊆ f −1 [f (X)] .<br />
Remark: <strong>The</strong> equality may not hold, for example, let f (x) = x 2 on R,<br />
and let X = [0, ∞), we have<br />
f −1 [f (X)] = f −1 [[0, ∞)] = R.<br />
(b) f (f −1 (Y )) ⊆ Y<br />
Proof: Given y ∈ f (f −1 (Y )) , then there exists a point x ∈ f −1 (Y )<br />
such that f (x) = y. Since x ∈ f −1 (Y ) , we know that f (x) ∈ Y. Hence,<br />
y ∈ Y. So, f (f −1 (Y )) ⊆ Y<br />
Remark: <strong>The</strong> equality may not hold, for example, let f (x) = x 2 on R,<br />
and let Y = R, we have<br />
f ( f −1 (Y ) ) = f (R) = [0, ∞) ⊆ R.<br />
(c) f −1 [Y 1 ∪ Y 2 ] = f −1 (Y 1 ) ∪ f −1 (Y 2 )<br />
Proof: Given x ∈ f −1 [Y 1 ∪ Y 2 ] , then f (x) ∈ Y 1 ∪ Y 2 . We consider two<br />
cases as follows.<br />
If f (x) ∈ Y 1 , then x ∈ f −1 (Y 1 ) . So, x ∈ f −1 (Y 1 ) ∪ f −1 (Y 2 ) .<br />
If f (x) /∈ Y 1 , i.e., f (x) ∈ Y 2 , then x ∈ f −1 (Y 2 ) . So, x ∈ f −1 (Y 1 ) ∪<br />
f −1 (Y 2 ) .<br />
From above, we have proved that<br />
f −1 [Y 1 ∪ Y 2 ] ⊆ f −1 (Y 1 ) ∪ f −1 (Y 2 ) . (*)<br />
Conversely, since f −1 (Y 1 ) ⊆ f −1 [Y 1 ∪ Y 2 ] and f −1 (Y 2 ) ⊆ f −1 [Y 1 ∪ Y 2 ] ,<br />
we have<br />
f −1 (Y 1 ) ∪ f −1 (Y 2 ) ⊆ f −1 [Y 1 ∪ Y 2 ] . (**)<br />
From (*) and (**), we have proved it.<br />
(d) f −1 [Y 1 ∩ Y 2 ] = f −1 (Y 1 ) ∩ f −1 (Y 2 )<br />
Proof: Given x ∈ f −1 (Y 1 ) ∩ f −1 (Y 2 ) , then f (x) ∈ Y 1 and f (x) ∈ Y 2 .<br />
So, f (x) ∈ Y 1 ∩ Y 2 . Hence, x ∈ f −1 [Y 1 ∩ Y 2 ] . That is, we have proved that<br />
f −1 (Y 1 ) ∩ f −1 (Y 2 ) ⊆ f −1 [Y 1 ∩ Y 2 ] . (*)<br />
9
Conversely, since f −1 [Y 1 ∩ Y 2 ] ⊆ f −1 (Y 1 ) and f −1 [Y 1 ∩ Y 2 ] ⊆ f −1 (Y 2 ) ,<br />
we have<br />
f −1 [Y 1 ∩ Y 2 ] ⊆ f −1 (Y 1 ) ∩ f −1 (Y 2 ) . (**)<br />
From (*) and (**), we have proved it.<br />
(e) f −1 (T − Y ) = S − f −1 (Y )<br />
Proof: Given x ∈ f −1 (T − Y ) , then f (x) ∈ T − Y. So, f (x) /∈ Y. We<br />
want to show that x ∈ S − f −1 (Y ) . Suppose NOT, then x ∈ f −1 (Y ) which<br />
implies that f (x) ∈ Y. That is impossible. Hence, x ∈ S − f −1 (Y ) . So, we<br />
have<br />
f −1 (T − Y ) ⊆ S − f −1 (Y ) . (*)<br />
Conversely, given x ∈ S −f −1 (Y ) , then x /∈ f −1 (Y ) . So, f (x) /∈ Y. That<br />
is, f (x) ∈ T − Y. Hence, x ∈ f −1 (T − Y ) . So, we have<br />
From (*) and (**), we have proved it.<br />
S − f −1 (Y ) ⊆ f −1 (T − Y ) . (**)<br />
(f) Generalize (c) and (d) to arbitrary unions and intersections.<br />
Proof: We give the statement without proof since it is the same as (c)<br />
and (d). In general, we have<br />
and<br />
f −1 (∪ i∈I A i ) = ∪ i∈I f −1 (A i ) .<br />
f −1 (∩ i∈I A i ) = ∩ i∈I f −1 (A i ) .<br />
Remark: From above sayings and Exercise 2.6, we found that the<br />
inverse image f −1 and the operations of sets, such as intersection and union,<br />
can be exchanged. However, for a function, we only have the exchange of<br />
f and the operation of union. <strong>The</strong> reader also see the Exercise 2.9 to get<br />
more.<br />
2.8 Refer to Exercise 2.7. Prove that f [f −1 (Y )] = Y for every subset Y<br />
of T if, and only if, T = f (S) .<br />
Proof: (⇒) It is clear that f (S) ⊆ T. In order to show the equality, it<br />
suffices to show that T ⊆ f (S) . Consider f −1 (T ) ⊆ S, then we have<br />
f ( f −1 (T ) ) ⊆ f (S) .<br />
10
By hyppothesis, we get T ⊆ f (S) .<br />
(⇐) Suppose NOT, i.e., f [f −1 (Y )] is a proper subset of Y for some<br />
Y ⊆ T by Exercise 2.7 (b). Hence, there is a y ∈ Y such that y /∈<br />
f [f −1 (Y )] . Since Y ⊆ f (S) = T, f (x) = y for some x ∈ S. It implies that<br />
x ∈ f −1 (Y ) . So, f (x) ∈ f [f −1 (Y )] which is impossible by the choice of y.<br />
Hence, f [f −1 (Y )] = Y for every subset Y of T.<br />
2.9 Let f : S → T be a function. Prove that the following statements<br />
are equivalent.<br />
(a) f is one-to-one on S.<br />
(b) f (A ∩ B) = f (A) ∩ f (B) for all subsets A, B of S.<br />
(c) f −1 [f (A)] = A for every subset A of S.<br />
(d) For all disjoint subsets A and B of S, the image f (A) and f (B) are<br />
disjoint.<br />
(e) For all subsets A and B of S with B ⊆ A, we have<br />
f (A − B) = f (A) − f (B) .<br />
Proof: (a) ⇒ (b) : Suppose that f is 1-1 on S. By Exercise 2.6, we<br />
have proved that f (A ∩ B) ⊆ f (A) ∩ f (B) for all A, B of S. In order to<br />
show the equality, it suffices to show that f (A) ∩ f (B) ⊆ f (A ∩ B) .<br />
Given y ∈ f (A) ∩ f (B) , then y = f (a) and y = f (b) where a ∈ A<br />
and b ∈ B. Since f is 1-1, we have a = b. That is, y ∈ f (A ∩ B) . So,<br />
f (A) ∩ f (B) ⊆ f (A ∩ B) .<br />
(b) ⇒ (c) : Suppose that f (A ∩ B) = f (A) ∩ f (B) for all subsets A, B<br />
of S. If A ≠ f −1 [f (A)] for some A of S, then by Exercise 2.7 (a), there is<br />
an element a /∈ A and a ∈ f −1 [f (A)] . Consider<br />
φ = f (A ∩ {a}) = f (A) ∩ f ({a}) by (b) (*)<br />
Since a ∈ f −1 [f (A)] , we have f (a) ∈ f (A) which contradicts to (*). Hence,<br />
no such a exists. That is, f −1 [f (A)] = A for every subset A of S.<br />
(c) ⇒ (d) : Suppose that f −1 [f (A)] = A for every subset A of S. If<br />
A ∩ B = φ, then Consider<br />
φ = A ∩ B<br />
= f −1 [f (A)] ∩ f −1 [f (B)]<br />
= f −1 (f (A) ∩ f (B)) by Exercise 2.7 (d)<br />
11
which implies that f (A) ∩ f (B) = φ.<br />
(d) ⇒ (e) : Suppose that for all disjoint subsets A and B of S, the image<br />
f (A) and f (B) are disjoint. If B ⊆ A, then since (A − B) ∩ B = φ, we have<br />
which implies that<br />
f (A − B) ∩ f (B) = φ<br />
f (A − B) ⊆ f (A) − f (B) . (**)<br />
Conversely, we consider if y ∈ f (A) − f (B) , then y = f (x) , where x ∈ A<br />
and x /∈ B. It implies that x ∈ A − B. So, y = f (x) ∈ f (A − B) . That is,<br />
By (**) and (***), we have proved it.<br />
f (A) − f (B) ⊆ f (A − B) . (***)<br />
(d) ⇒ (a) : Suppose that f (A − B) = f (A) − f (B) for all subsets A and<br />
B of S with B ⊆ A. If f (a) = f (b) , consider A = {a, b} and B = {b} , we<br />
have<br />
f (A − B) = φ<br />
which implies that A = B. That is, a = b. So, f is 1-1.<br />
2.10 Prove that if A˜B and B˜C, then A˜C.<br />
Proof: Since A˜B and B˜C, then there exists bijection f and g such<br />
that<br />
f : A → B and g : B → C.<br />
So, if we consider g ◦ f : A → C, then A˜C since g ◦ f is bijection.<br />
2.11 If {1, 2, ..., n} ˜ {1, 2, ..., m} , prove that m = n.<br />
Proof: Since {1, 2, ..., n} ˜ {1, 2, ..., m} , there exists a bijection function<br />
f : {1, 2, ..., n} → {1, 2, ..., m} .<br />
Since f is 1-1, then n ≤ m. Conversely, consider f −1 is 1-1 to get m ≤ n. So,<br />
m = n.<br />
2.12 If S is an infinite set, prove that S contains a countably infinite<br />
subset. Hint. Choose an element a 1 in S and consider S − {a 1 } .<br />
12
Proof: Since S is an infinite set, then choose a 1 in S and thus S − {a 1 }<br />
is still infinite. From this, we have S − {a 1 , .., a n } is infinite. So, we finally<br />
have<br />
{a 1 , ..., a n , ...} (⊆ S)<br />
which is countably infinite subset.<br />
2.13 Prove that every infinite set S contains a proper subset similar to S.<br />
Proof: By Exercise 2.12, we write S = ˜S ∪ {x 1 , ..., x n , ...} , where ˜S ∩<br />
{x 1 , ..., x n , ...} = φ and try to show<br />
as follows. Define<br />
by<br />
˜S ∪ {x 2 , ..., x n , ...} ˜S<br />
f : ˜S ∪ {x 2 , ..., x n , ...} → S = ˜S ∪ {x 1 , ..., x n , ...}<br />
f (x) =<br />
{<br />
x if x ∈ ˜S<br />
x i if x = x i+1<br />
.<br />
<strong>The</strong>n it is clear that f is 1-1 and onto. So, we have proved that every infinite<br />
set S contains a proper subset similar to S.<br />
Remark: In the proof, we may choose the map<br />
f : ˜S ∪ {x N+1 , ..., x n , ...} → S = ˜S ∪ {x 1 , ..., x n , ...}<br />
by<br />
f (x) =<br />
{<br />
x if x ∈ ˜S<br />
x i if x = x i+N<br />
.<br />
2.14 If A is a countable set and B an uncountable set, prove that B − A<br />
is similar to B.<br />
Proof: In order to show it, we consider some cases as follows. (i) B∩A =<br />
φ (ii) B ∩ A is a finite set, and (iii) B ∩ A is an infinite set.<br />
For case (i), B − A = B. So, B − A is similar to B.<br />
For case (ii), it follows from the Remark in Exercise 2.13.<br />
For case (iii), note that B ∩ A is countable, and let C = B − A, we have<br />
B = C ∪ (B ∩ A) . We want to show that<br />
(B − A) ˜B ⇔ C˜C ∪ (B ∩ A) .<br />
13
By Exercise 2.12, we write C = ˜C ∪ D, where D is countably infinite and<br />
˜C ∩ D = φ. Hence,<br />
( )<br />
˜C ∪ D<br />
C˜C ∪ (B ∩ A) ⇔<br />
⇔<br />
(<br />
˜C ∪ D<br />
)<br />
˜<br />
[ ]<br />
˜ ˜C ∪ (D ∪ (B ∩ A))<br />
(<br />
˜C ∪ D<br />
′)<br />
( ) ( )<br />
where D ′ = D∪(B ∩ A) which is countably infinite. Since ˜C ∪ D ˜ ˜C ∪ D<br />
′<br />
is clear, we have proved it.<br />
2.15 A real number is called algebraic if it is a root of an algebraic<br />
equation f (x) = 0, where f (x) = a 0 + a 1 x + ... + a n x n is a polynomial<br />
with integer coefficients. Prove that the set of all polynomials with integer<br />
coefficients is countable and deduce that the set of algebraic numbers is also<br />
countable.<br />
Proof: Given a positive integer N (≥ 2) , there are only finitely many<br />
eqautions with<br />
n∑<br />
n + |a k | = N, where a k ∈ Z. (*)<br />
k=1<br />
Let S N = {f : f (x) = a 0 + a 1 x + ... + a n x n satisfies (*)} , then S N is a finite<br />
set. Hence, ∪ ∞ n=2S n is countable. Clearly, the set of all polynomials with<br />
integer coefficients is a subset of ∪ ∞ n=2S n . So, the set of all polynomials with<br />
integer coefficients is countable. In addition, a polynomial of degree k has at<br />
most k roots. Hence, the set of algebraic numbers is also countable.<br />
2.16 Let S be a finite set consisting of n elements and let T be the<br />
collection of all subsets of S. Show that T is a finite set and find the number<br />
of elements in T.<br />
Proof: Write S = {x 1 , ..., x n } , then T =the collection of all subsets of<br />
S. Clearly, T is a finite set with 2 n elements.<br />
2.17 Let R denote the set of real numbers and let S denote the set<br />
of all real-valued functions whose domain in R. Show that S and R are not<br />
equinumrous. Hint. Assume S˜R and let f be a one-to-one function such<br />
that f (R) = S. If a ∈ R, let g a = f (a) be the real-valued function in S which<br />
correspouds to real number a. Now define h by the equation h (x) = 1+g x (x)<br />
if x ∈ R, and show that h /∈ S.<br />
14
Proof: Assume S˜R and let f be a one-to-one function such that f (R) =<br />
S. If a ∈ R, let g a = f (a) be the real-valued function in S which correspouds<br />
to real number a. Define h by the equation h (x) = 1 + g x (x) if x ∈ R, then<br />
which implies that<br />
h = f (b) = g b<br />
h (b) := 1 + g b (b) = g b (b)<br />
which is impossible. So, S and R are not equinumrous.<br />
Remark: <strong>The</strong>re is a similar exercise, we write it as a reference. <strong>The</strong><br />
cardinal number of C [a, b] is 2 ℵ 0<br />
, where ℵ 0 = # (N) .<br />
Proof: First, # (R) = 2 ℵ 0<br />
≤ # (C [a, b]) by considering constant function.<br />
Second, we consider the map<br />
by<br />
f : C [a, b] → P (Q × Q) , the power set of Q × Q<br />
f (ϕ) = {(x, y) ∈ Q × Q : x ∈ [a, b] and y ≤ ϕ (x)} .<br />
Clearly, f is 1-1. It implies that # (C [a, b]) ≤ # (P (Q × Q)) = 2 ℵ 0<br />
.<br />
So, we have proved that # (C [a, b]) = 2 ℵ 0<br />
.<br />
Note: For notations, the reader can see the textbook, in Chapter 4, pp<br />
102. Also, see the book, Set <strong>The</strong>ory and Related Topics by Seymour<br />
Lipschutz, Chapter 9, pp 157-174. (Chinese Version)<br />
2.18 Let S be the collection of all sequences whose terms are the integers<br />
0 and 1. Show that S is uncountable.<br />
Proof: Let E be a countable subet of S, and let E consists of the sequences<br />
s 1 , .., s n , .... We construct a sequence s as follows. If the nth digit<br />
in s n is 1, we let the nth digit of s be 0, and vice versa. <strong>The</strong>n the sequence<br />
s differes from every member of E in at least one place; hence s /∈ E. But<br />
clearly s ∈ S, so that E is a proper subset of S.<br />
We have shown that every countable subset of S is a proper subset of S.<br />
It follows that S is uncountable (for otherwise S would be a proper subset<br />
of S, which is absurb).<br />
Remark: In this exercise, we have proved that R, the set of real numbers,<br />
is uncountable. It can be regarded as the Exercise 1.22 for k = 2. (Binary<br />
System).<br />
15
2.19 Show that the following sets are countable:<br />
(a) the set of circles in the complex plane having the ratiional radii and<br />
centers with rational coordinates.<br />
Proof: Write the set of circles in the complex plane having the ratiional<br />
radii and centers with rational coordinates, {C (x n ; q n ) : x n ∈ Q × Q and q n ∈ Q} :=<br />
S. Clearly, S is countable.<br />
(b) any collection of disjoint intervals of positive length.<br />
Proof: Write the collection of disjoint intervals of positive length, {I : I is an interval of positive<br />
S. Use the reason in Exercise 2.21, we have proved that S is countable.<br />
2.20 Let f be a real-valued function defined for every x in the interval<br />
0 ≤ x ≤ 1. Suppose there is a positive number M having the following<br />
property: for every choice of a finite number of points x 1 , x 2 , ..., x n in the<br />
interval 0 ≤ x ≤ 1, the sum<br />
|f (x 1 ) + ... + f (x n )| ≤ M.<br />
Let S be the set of those x in 0 ≤ x ≤ 1 for which f (x) ≠ 0. Prove that S<br />
is countable.<br />
Proof: Let S n = {x ∈ [0, 1] : |f (x)| ≥ 1/n} , then S n is a finite set by<br />
hypothesis. In addition, S = ∪ ∞ n=1S n . So, S is countable.<br />
2.21 Find the fallacy in the following ”proof” that the set of all intervals<br />
of positive length is countable.<br />
Let {x 1 , x 2 , ...} denote the countable set of rational numbers and let I<br />
be any interval of positive length. <strong>The</strong>n I contains infinitely many rational<br />
points x n , but among these there will be one with smallest index n. Define<br />
a function F by means of the eqaution F (I) = n if x n is the rational number<br />
with smallest index in the interval I. This function establishes a one-to-one<br />
correspondence between the set of all intervals and a subset of the positive<br />
integers. Hence, the set of all intervals is countable.<br />
Proof: Note that F is not a one-to-one correspondence between the set<br />
of all intervals and a subset of the positive integers. So, this is not a proof.<br />
In fact, the set of all intervals of positive length is uncountable.<br />
Remark: Compare with Exercise 2.19, and the set of all intervals of<br />
positive length is uncountable is clear by considering {(0, x) : 0 < x < 1} .<br />
16
2.22 Let S denote the collection of all subsets of a given set T. Let f : S →<br />
R be a real-valued function defined on S. <strong>The</strong> function f is called additive<br />
if f (A ∪ B) = f (A) + f (B) whenever A and B are disjoint subsets of T. If<br />
f is additive, prove that for any two subsets A and B we have<br />
f (A ∪ B) = f (A) + f (B − A)<br />
and<br />
f (A ∪ B) = f (A) + f (B) − f (A ∩ B) .<br />
Proof: Since A ∩ (B − A) = φ and A ∪ B = A ∪ (B − A) , we have<br />
f (A ∪ B) = f (A ∪ (B − A)) = f (A) + f (B − A) . (*)<br />
In addition, since(B − A) ∩ (A ∩ B) = φ and B = (B − A) ∪ (A ∩ B) , we<br />
have<br />
f (B) = f ((B − A) ∪ (A ∩ B)) = f (B − A) + f (A ∩ B)<br />
which implies that<br />
By (*) and (**), we have proved that<br />
f (B − A) = f (B) − f (A ∩ B) (**)<br />
f (A ∪ B) = f (A) + f (B) − f (A ∩ B) .<br />
2.23 Refer to Exercise 2.22. Assume f is additive and assume also that<br />
the following relations hold for two particular subsets A and B of T :<br />
and<br />
and<br />
f (A ∪ B) = f (A ′ ) + f (B ′ ) − f (A ′ ) f (B ′ )<br />
f (A ∩ B) = f (A) f (B)<br />
f (A) + f (B) ≠ f (T ) ,<br />
where A ′ = T − A, B ′ = T − B. Prove that these relations determine f (T ) ,<br />
and compute the value of f (T ) .<br />
17
Proof: Write<br />
f (T ) = f (A) + f (A ′ ) = f (B) + f (B ′ ) ,<br />
then<br />
[f (T )] 2 = [f (A) + f (A ′ )] [f (B) + f (B ′ )]<br />
= f (A) f (B) + f (A) f (B ′ ) + f (A ′ ) f (B) + f (A ′ ) f (B ′ )<br />
= f (A) f (B) + f (A) [f (T ) − f (B)] + [f (T ) − f (A)] f (B) + f (A ′ ) f (B ′ )<br />
= [f (A) + f (B)] f (T ) − f (A) f (B) + f (A ′ ) f (B ′ )<br />
= [f (A) + f (B)] f (T ) − f (A) f (B) + f (A ′ ) + f (B ′ ) − f (A ∪ B)<br />
= [f (A) + f (B)] f (T ) − f (A) f (B) + [f (T ) − f (A)] + [f (T ) − f (B)]<br />
− [f (A) + f (B) − f (A ∩ B)]<br />
= [f (A) + f (B) + 2] f (T ) − f (A) f (B) − 2 [f (A) + f (B)] + f (A ∩ B)<br />
= [f (A) + f (B) + 2] f (T ) − 2 [f (A) + f (B)]<br />
which implies that<br />
[f (T )] 2 − [f (A) + f (B) + 2] f (T ) + 2 [f (A) + f (B)] = 0<br />
which implies that<br />
x 2 − (a + 2) x + 2a = 0 ⇒ (x − a) (x − 2) = 0<br />
where a = f (A) + f (B) . So, x = 2 since x ≠ a by f (A) + f (B) ≠ f (T ) .<br />
18
Charpter 3 Elements of Point set Topology<br />
Open and closed sets in R 1 and R 2<br />
3.1 Prove that an open interval in R 1 is an open set and that a closed interval is a<br />
closed set.<br />
proof: 1. Let a, b be an open interval in R 1 , and let x a, b. Consider<br />
minx a, b x : L. <strong>The</strong>n we have Bx, L x L, x L a, b. Thatis,x is an<br />
interior point of a, b. Sincex is arbitrary, we have every point of a, b is interior. So,<br />
a, b is open in R 1 .<br />
2. Let a, b be a closed interval in R 1 , and let x be an adherent point of a, b. Wewant<br />
to show x a, b. Ifx a, b, then we have x a or x b. Consider x a, then<br />
Bx, a 2<br />
x a, b 3x 2<br />
a , x 2<br />
a a, b <br />
which contradicts the definition of an adherent point. Similarly for x b.<br />
<strong>The</strong>refore, we have x a, b if x is an adherent point of a, b. Thatis,a, b contains<br />
its all adherent points. It implies that a, b is closed in R 1 .<br />
3.2 Determine all the accumulation points of the following sets in R 1 and decide<br />
whether the sets are open or closed (or neither).<br />
(a) All integers.<br />
Solution: Denote the set of all integers by Z. Letx Z, and consider<br />
Bx, x1 x S . So,Z has no accumulation points.<br />
2<br />
However, Bx, x1 S x . SoZ contains its all adherent points. It means that<br />
2<br />
Z is closed. Trivially, Z is not open since Bx, r is not contained in Z for all r 0.<br />
Remark: 1. Definition of an adherent point: Let S be a subset of R n ,andx a point in<br />
R n , x is not necessarily in S. <strong>The</strong>n x is said to be adherent to S if every n ball Bx<br />
contains at least one point of S. To be roughly, Bx S .<br />
2. Definition of an accumulation point: Let S be a subset of R n ,andx a point in R n ,<br />
then x is called an accumulation point of S if every n ball Bx contains at least one point<br />
of S distinct from x. To be roughly, Bx x S . Thatis,x is an accumulation<br />
point if, and only if, x adheres to S x. Note that in this sense,<br />
Bx x S Bx S x.<br />
3. Definition of an isolated point: If x S, but x is not an accumulation point of S, then<br />
x is called an isolated point.<br />
4. Another solution for Z is closed: Since R Z nZ n, n 1, we know that R Z<br />
is open. So, Z is closed.<br />
5. In logics, if there does not exist any accumulation point of a set S, then S is<br />
automatically a closed set.<br />
(b) <strong>The</strong> interval a, b.<br />
solution: In order to find all accumulation points of a, b, we consider 2 cases as<br />
follows.<br />
1. a, b :Letx a, b, then Bx, r x a, b for any r 0. So, every<br />
point of a, b is an accumulation point.<br />
2. R 1 a, b , a b, : For points in b, and , a, it is easy to know<br />
that these points cannot be accumulation points since x b, or x , a, there
exists an n ball Bx, r x such that Bx, r x x a, b . For the point a, it is easy<br />
to know that Ba, r a a, b . That is, in this case, there is only one<br />
accumulation point a of a, b.<br />
So, from 1 and 2, we know that the set of the accumulation points of a, b is a, b.<br />
Since a a, b, we know that a, b cannot contain its all accumulation points. So,<br />
a, b is not closed.<br />
Since an n ball Bb, r is not contained in a, b for any r 0, we know that the point<br />
b is not interior to a, b. So,a, b is not open.<br />
(c) All numbers of the form 1/n, (n 1,2,3,....<br />
Solution: Write the set 1/n : n 1,2,... 1, 1/2, 1/3, . . . , 1/n,... : S.<br />
Obviously, 0 is the only one accumulation point of S. So,S is not closed since S does not<br />
contain the accumulation point 0. Since 1 S, andB1, r is not contained in S for any<br />
r 0, S is not open.<br />
Remark: Every point of 1/n : n 1,2,3,... is isolated.<br />
(d) All rational numbers.<br />
Solutions: Denote all rational numbers by Q. It is trivially seen that the set of<br />
accumulation points is R 1 .<br />
So, Q is not closed. Consider x Q, anyn ball Bx is not contained in Q. Thatis,x<br />
is not an interior point of Q. In fact, every point of Q is not an interior point of Q. So,<br />
Q is not open.<br />
(e) All numbers of the form 2 n 5 m , (m, n 1,2,....<br />
Solution: Write the set<br />
2 n 5 m : m, n 1, 2, . . m m1 1 2 5m , 1 4 5m ,..., 1<br />
2<br />
n 5m ,... : S<br />
1 2 1 5 , 1 2 1 5 ,..., 1 2 2 5 1<br />
m ,... <br />
1 4 1 5 , 1 4 1 5 ,..., 1 2 4 5 1<br />
m ,... <br />
........................................................<br />
<br />
2 1 n 1 5 , 1<br />
2<br />
n 1 5 ,..., 1 2 2<br />
n 5 1<br />
m ... <br />
.....................................................<br />
So,wefindthatS 1 : n 1, 2, . . . 1 : m 1, 2, . . .<br />
2 n 5<br />
0. So,S is not<br />
m<br />
closed since it does not contain 0. Since 1 S, andB 1 , r is not contained in S for any<br />
2<br />
2<br />
r 0, S is not open.<br />
Remark: By (1)-(3), we can regard them as three sequences<br />
1<br />
2 m 5m , 1 5 m m<br />
and 1<br />
m1 4 m1 2<br />
n m 5m , respectively.<br />
m1<br />
<strong>And</strong> it means that for (1), the sequence 5 m m m1<br />
moves 1 . Similarly for others. So, it is<br />
2<br />
easy to see why 1 is an accumulation point of 1 2 2 5m m<br />
m1 . <strong>And</strong> thus get the set of all<br />
accumulation points of 2 n 5 m : m, n 1, 2, . . .<br />
(f) All numbers of the form 1 n 1/m, (m, n 1,2,....<br />
Solution: Write the set of all numbers 1 n 1/m, (m, n 1, 2, . . . as<br />
1 m 1 m m<br />
1 1<br />
m1 m : S.<br />
m1<br />
1<br />
2<br />
3
<strong>And</strong> thus by the remark in (e), it is easy to know that S 1, 1. So,S is not closed<br />
since S S. Since2 S, andB2, r is not contained in S for any r 0, S is not open.<br />
(g) All numbers of the form 1/n 1/m, (m, n 1,2,....<br />
Solution: Write the set of all numbers 1/n 1/m, (m, n 1, 2, . . . as<br />
1 1/m m m1<br />
1/2 1/m m m1<br />
...1/n 1/m m m1<br />
...: S.<br />
We find that S 1/n : n N 1/m : m N 0 1/n : n N 0. So,S is<br />
not closed since S S. Since1 S, andB1, r is not contained in S for any r 0, S is<br />
not open.<br />
(h) All numbers of the form 1 n /1 1/n, (n 1,2,....<br />
Soluton: Write the set of all numbers 1 n /1 1/n, (n 1, 2, . . . as<br />
k<br />
1<br />
1 1 2k k1<br />
<br />
1<br />
1 1<br />
2k1<br />
k<br />
k1<br />
: S.<br />
We find that S 1, 1. So,S is not closed since S S. Since 1 S, andB 1<br />
2 2<br />
not contained in S for any r 0, S is not open.<br />
, r is<br />
3.3 <strong>The</strong> same as Exercise 3.2 for the following sets in R 2 .<br />
(a) All complex z such that |z| 1.<br />
Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1.<br />
So, S is not closed since S S. Letz S, then |z| 1. Consider Bz, |z|1 S, soevery<br />
2<br />
point of S is interior. That is, S is open.<br />
(b) All complex z such that |z| 1.<br />
Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1.<br />
So, S is closed since S S. Since1 S, andB1, r is not contained in S for any r 0, S<br />
is not open.<br />
(c) All complex numbers of the form 1/n i/m, (m, n 1, 2, . . . .<br />
Solution: Write the set of all complex numbers of the form 1/n i/m,<br />
(m, n 1, 2, . . . as<br />
1 m i m<br />
1 i m<br />
m1 2 m ... 1<br />
m1<br />
n m i m<br />
...: S.<br />
m1<br />
We know that S 1/n : n 1,2,... i/m : m 1, 2, . . . 0. So,S is not closed<br />
since S S. Since1 i S, andB1 i, r is not contained in S for any r 0, S is not<br />
open.<br />
(d) All points x, y such that x 2 y 2 1.<br />
Solution: Denote x, y : x 2 y 2 1 by S. We know that<br />
S x, y : x 2 y 2 1. So,S is not closed since S S. Letp x, y S, then<br />
x 2 y 2 1. Itiseasytofindthatr 0 such that Bp, r S. So,S is open.<br />
(e) All points x, y such that x 0.<br />
Solution: Write all points x, y such that x 0asx, y : x 0 : S. It is easy to<br />
know that S x, y : x 0. So,S is not closed since S S. Letx S, then it is easy<br />
to find r x 0 such that Bx, r x S. So,S is open.<br />
(f) All points x, y such that x 0.<br />
Solution: Write all points x, y such that x 0asx, y : x 0 : S. It is easy to
know that S x, y : x 0. So,S is closed since S S. Since0, 0 S, and<br />
B0, 0, r is not contained in S for any r 0, S is not open.<br />
3.4 Prove that every nonempty open set S in R 1 contains both rational and irratonal<br />
numbers.<br />
proof: Given a nonempty open set S in R 1 .Letx S, then there exists r 0 such that<br />
Bx, r S since S is open. <strong>And</strong> in R 1 , the open ball Bx, r x r, x r. Since any<br />
interval contains both rational and irrational numbers, we have S contains both rational and<br />
irrational numbers.<br />
3.5 Prove that the only set in R 1 which are both open and closed are the empty set<br />
and R 1 itself. Is a similar statement true for R 2 <br />
Proof: Let S be the set in R 1 , and thus consider its complement T R 1 S. <strong>The</strong>n we<br />
have both S and T are open and closed. Suppose that S R 1 and S , we will show that<br />
it is impossible as follows.<br />
Since S R 1 ,andS , then T and T R 1 . Choose s 0 S and t 0 T, then we<br />
consider the new point s 0t 0<br />
which is in S or T since R S T. If s 0t 0<br />
S, wesay<br />
2 2<br />
s 0 t 0<br />
s<br />
2 1, otherwise, we say s 0t 0<br />
t<br />
2 1.<br />
Continue these steps, we finally have two sequences named s n S and t m T.<br />
In addition, the two sequences are convergent to the same point, say p by our construction.<br />
So, we get p S and p T since both S and T are closed.<br />
However, it leads us to get a contradiction since p S T . Hence S R 1 or<br />
S .<br />
Remark: 1. In the proof, the statement is true for R n .<br />
2. <strong>The</strong> construction is not strange for us since the process is called Bolzano Process.<br />
3. 6 Prove that every closed set in R 1 is the intersection of a countable collection of<br />
open sets.<br />
proof: Given a closed set S, and consider its complement R 1 S which is open. If<br />
R 1 S , there is nothing to prove. So, we can assume that R 1 S .<br />
Let x R 1 S, then x is an interior point of R 1 S. So, there exists an open interval<br />
a, b such that x a, b R 1 S. In order to show our statement, we choose a smaller<br />
interval a x , b x so that x a x , b x and a x , b x a, b R 1 S. Hence, we have<br />
R 1 S xR 1 S a x , b x <br />
which implies that<br />
S R 1 xR 1 S a x , b x <br />
xR 1 S R 1 a x , b x <br />
n n1 R 1 a n , b n (by Lindelof Convering <strong>The</strong>orem).<br />
Remark: 1. <strong>The</strong>re exists another proof by Representation <strong>The</strong>orem for Open Sets on<br />
<strong>The</strong> <strong>Real</strong> Line.<br />
2. Note that it is true for that every closed set in R 1 is the intersection of a countable<br />
collection of closed sets.<br />
3. <strong>The</strong> proof is suitable for R n if the statement is that every closed set in R n is the<br />
intersection of a countable collection of open sets. All we need is to change intervals into<br />
disks.
3.7 Prove that a nonempty, bounded closed set S in R 1 is either a closed interval, or<br />
that S can be obtained from a closed interval by removing a countable disjoint collection of<br />
open intervals whose endpoints belong to S.<br />
proof: If S is an interval, then it is clear that S is a closed interval. Suppose that S is not<br />
an interval. Since S is bounded and closed, both sup S and inf S are in S. So,R 1 S<br />
inf S,supS S. Denote inf S,supS by I. Consider R 1 S is open, then by<br />
Representation <strong>The</strong>orem for Open Sets on <strong>The</strong> <strong>Real</strong> Line, we have<br />
R 1 S m m1 I m<br />
I S<br />
which implies that<br />
S I m m1 I m .<br />
That is, S can be obtained from a closed interval by removing a countable disjoint<br />
collection of open intervals whose endpoints belong to S.<br />
Open and closed sets in R n<br />
3.8 Prove that open n balls and n dimensional open intervals are open sets in R n .<br />
proof: Given an open n ball Bx, r. Choose y Bx, r and thus consider<br />
By, d Bx, r, whered min|x y|, r |x y|. <strong>The</strong>n y is an interior point of Bx, r.<br />
Since y is arbitrary, we have all points of Bx, r are interior. So, the open n ball Bx, r is<br />
open.<br />
Given an n dimensional open interval a 1 , b 1 a 2 , b 2 ...a n , b n : I. Choose<br />
x x 1, x 2 ,...,x n I and thus consider r minin i1 r i ,wherer i minx i a i , b i x i .<br />
<strong>The</strong>n Bx, r I. Thatis,x is an interior point of I. Sincex is arbitrary, we have all points<br />
of I are interior. So, the n dimensional open interval I is open.<br />
3.9 Prove that the interior of a set in R n isopeninR n .<br />
Proof: Let x intS, then there exists r 0 such that Bx, r S. Choose any point of<br />
Bx, r, sayy. <strong>The</strong>n y is an interior point of Bx, r since Bx, r is open. So, there exists<br />
d 0 such that By, d Bx, r S. Soy is also an interior point of S. Sincey is<br />
arbitrary, we find that every point of Bx, r is interior to S. Thatis,Bx, r intS. Sincex<br />
is arbitrary, we have all points of intS are interior. So, intS is open.<br />
Remark: 1 It should be noted that S is open if, and only if S intS.<br />
2. intintS intS.<br />
3. If S T, then intS intT.<br />
3.10 If S R n , prove that intS is the union of all open subsets of R n which are<br />
contained in S. This is described by saying that intS is the largest open subset of S.<br />
proof: It suffices to show that intS AS A,whereA is open. To show the statement,<br />
we consider two steps as follows.<br />
1. Let x intS, then there exists r 0 such that Bx, r S. So,<br />
x Bx, r AS A.Thatis,intS AS A.<br />
2. Let x AS A, then x A for some open set A S. SinceA is open, x is an<br />
interior point of A. <strong>The</strong>re exists r 0 such that Bx, r A S. Sox is an interior point<br />
of S, i.e., x intS. Thatis, AS A intS.<br />
From 1 and 2, we know that intS AS A,whereA is open.<br />
Let T be an open subset of S such that intS T. SinceintS AS A,whereA is open,
we have intS T AS A which implies intS T by intS AS A. Hence, intS is the<br />
largest open subset of S.<br />
3.11 If S and T are subsets of R n , prove that<br />
intS intT intS T and intS intT intS T.<br />
Proof: For the part intS intT intS T, we consider two steps as follows.<br />
1. Since intS S and intT T, wehaveintS intT S T which implies<br />
that Note that intS intT is open.<br />
intS intT intintS intT intS T.<br />
2. Since S T S and S T T, wehaveintS T intS and<br />
intS T intT. So,<br />
intS T intS intT.<br />
From 1 and 2, we know that intS intT intS T.<br />
For the part intS intT intS T, we consider intS S and intT T. So,<br />
intS intT S T<br />
which implies that Note that intS intT is open.<br />
intintS intT intS intT intS T.<br />
Remark: It is not necessary that intS intT intS T. For example, let S Q,<br />
and T Q c , then intS , andintT . However, intS T intR 1 R.<br />
3.12 Let S denote the derived set and S theclosureofasetSinR n . Prove that<br />
(a) S is closed in R n ; that is S S .<br />
proof: Let x be an adherent point of S . In order to show S is closed, it suffices to<br />
show that x is an accumulation point of S. Assume x is not an accumulation point of S, i.e.,<br />
there exists d 0 such that<br />
Bx, d x S . *<br />
Since x adheres to S , then Bx, d S . So, there exists y Bx, d such that y is an<br />
accumulation point of S. Note that x y, by assumption. Choose a smaller radius d so that<br />
By, d Bx, d x and By, d S .<br />
It implies<br />
By, d S Bx, d x S by (*)<br />
which is absurb. So, x is an accumulation point of S. Thatis,S contains all its adherent<br />
points. Hence S is closed.<br />
(b) If S T, then S T .<br />
Proof: Let x S , then Bx, r x S for any r 0. It implies that<br />
Bx, r x T for any r 0sinceS T. Hence, x is an accumulation point of T.<br />
That is, x T .So,S T .<br />
(c) S T S T <br />
Proof: For the part S T S T , we show it by two steps.<br />
1. Since S S T and T S T, wehaveS S T and T S T by (b).<br />
So,<br />
S T S T <br />
2. Let x S T , then Bx, r x S T . Thatis,
Bx, r x S Bx, r x T .<br />
So, at least one of Bx, r x S and Bx, r x T is not empty. If<br />
Bx, r x S , then x S .<strong>And</strong>ifBx, r x T , then x T .So,<br />
S T S T .<br />
From 1 and 2, we have S T S T .<br />
Remark: Note that since S T S T ,wehaveclS T clS clT,<br />
where clS is the closure of S.<br />
(d) S S .<br />
Proof: Since S S S , then S S S S S S since S S by<br />
(a).<br />
(e) S is closed in R n .<br />
Proof: Since S S S by (d), then S cantains all its accumulation points. Hence, S<br />
is closed.<br />
Remark: <strong>The</strong>re is another proof which is like (a). But it is too tedious to write.<br />
(f) S is the intersection of all closed subsets of R n containing S. That is, S is the<br />
smallest closed set containing S.<br />
Proof: It suffices to show that S AS A,whereA is closed. To show the statement,<br />
we consider two steps as follows.<br />
1. Since S is closed and S S, then AS A S.<br />
2. Let x S, then Bx, r S for any r 0. So, if A S, then<br />
Bx, r A for any r 0. It implies that x is an adherent point of A. Hence if A S,<br />
and A is closed, we have x A. Thatis,x AS A.So,S AS A.<br />
From 1 and 2, we have S AS A.<br />
Let S T S, whereT is closed. <strong>The</strong>n S AS A T. It leads us to get T S.<br />
That is, S is the smallest closed set containing S.<br />
Remark: In the exercise, there has something to remeber. We list them below.<br />
Remark 1. If S T, then S T .<br />
2. If S T, then S T.<br />
3. S S S .<br />
4. S is closed if, and only if S S.<br />
5. S is closed.<br />
6. S is the smallest closed set containing S.<br />
3.13 Let S and T be subsets of R n . Prove that clS T clS clT and that<br />
S clT clS T if S is open, where clS is the closure of S.<br />
Proof: Since S T S and S T T, then clS T clS and,<br />
clS T clT. So,clS T clS clT.<br />
Given an open set S , and let x S clT, then we have
1. x S and S is open.<br />
Bx, d S for some d 0.<br />
<br />
Bx, r S Bx, r if r d.<br />
Bx, r S Bx, d if r d.<br />
and<br />
2. x clT<br />
Bx, r T for any r 0.<br />
From 1 and 2, we know<br />
Bx, r S T Bx, r S T Bx, r T if r d.<br />
Bx, r S T Bx, r S T Bx, d T if r d.<br />
So, it means that x is an adherent point of S T. Thatis,x clS T. Hence,<br />
S clT clS T.<br />
Remark: It is not necessary that clS T clS clT. For example, S Q and<br />
T Q c , then clS T and clS clT R 1 .<br />
Note. <strong>The</strong> statements in Exercises 3.9 through 3.13 are true in any metric space.<br />
3.14 AsetSinR n is called convex if, for every pair of points x and y in S and every<br />
real satisfying 0 1, we have x 1 y S. Interpret this statement<br />
geometrically (in R 2 and R 3 and prove that<br />
(a) Every n ball in R n is convex.<br />
Proof: Given an n ball Bp, r, and let x, y Bp, r. Consider x 1 y, where<br />
0 1.<br />
<strong>The</strong>n<br />
x 1 y p x p 1 y p<br />
x p 1 y p<br />
r 1 r<br />
r.<br />
So, we have x 1 y Bp, r for 0 1. Hence, by the definition of convex,<br />
we know that every n ball in R n is convex.<br />
(b) Every n dimensional open interval is convex.<br />
Proof: Given an n dimensional open interval I a 1 , b 1 ...a n , b n .Letx, y I,<br />
and thus write x x 1 , x 2 ,...,x n and y y 1 , y 2 ,...y n . Consider<br />
x 1 y x 1 1 y 1 , x 2 1 y 2 ,...,x n 1 y n where 0 1.<br />
<strong>The</strong>n<br />
a i x i 1 y i b i ,wherei 1, 2, . . , n.<br />
So, we have x 1 y I for 0 1. Hence, by the definition of convex, we<br />
know that every n dimensional open interval is convex.<br />
(c) <strong>The</strong> interior of a convex is convex.<br />
Proof: Given a convex set S, and let x, y intS. <strong>The</strong>n there exists r 0 such that<br />
Bx, r S, andBy, r S. Consider x 1 y : p S, where 0 1, since S<br />
is convex.
Claim that Bp, r S as follows.<br />
Let q Bp, r, We want to find two special points x Bx, r, andy By, r such<br />
that q x 1 y.<br />
Since the three n balls Bx, r, By, r, andBp, r have the same radius. By<br />
parallelogram principle, we let x q x p, andy q y p, then<br />
x x q p r, andy y q p r.<br />
It implies that x Bx, r, andy By, r. In addition,<br />
x 1 y<br />
q x p 1 q y p<br />
q.<br />
Since x, y S, andS is convex, then q x 1 y S. It implies that Bp, r S<br />
since q is arbitrary. So, we have proved the claim. That is, for 0 1,<br />
x 1 y p intS if x, y intS, andS is convex. Hence, by the definition of<br />
convex, we know that the interior of a convex is convex.<br />
(d) <strong>The</strong> closure of a convex is convex.<br />
Proof: Given a convex set S, and let x, y S. Consider x 1 y : p, where<br />
0 1, and claim that p S, i.e., we want to show that Bp, r S .<br />
Suppose NOT, there exists r 0 such that<br />
Bp, r S . *<br />
Since x, y S, then Bx, r S and By, r S . <strong>And</strong> let x Bx, r S and<br />
2 2 2<br />
y By, r S. Consider<br />
2<br />
x 1 y p x 1 y x 1 y<br />
x x 1 y 1 y<br />
<br />
x x x x <br />
1 y 1 y 1 y 1 y<br />
x x 1 y y | |x y<br />
<br />
2 r | |x y<br />
r<br />
if we choose a suitable number , where 0 1.<br />
Hence, we have the point x 1 y Bp, r. Note that x, y S and S is convex,<br />
we have x 1 y S. It leads us to get a contradiction by (*). Hence, we have proved<br />
the claim. That is, for 0 1, x 1 y p S if x, y S. Hence, by the<br />
definition of convex, we know that the closure of a convex is convex.<br />
3.15 Let F be a collection of sets in R n , and let S AF A and T AF A. For each<br />
of the following statements, either give a proof or exhibit a counterexample.<br />
F.<br />
(a) If x is an accumulation point of T, then x is an accumulation point of each set A in<br />
Proof: Let x be an accumulation point of T, then Bx, r x T for any<br />
r 0. Note that for any A F, wehaveT A. Hence Bx, r x A for any<br />
r 0. That is, x is an accumulation point of A for any A F.<br />
<strong>The</strong> conclusion is that If x is an accumulation point of T AF A, then x is an<br />
accumulation point of each set A in F.
(b) If x is an accumulation point of S, then x is an accumulation point of at least one set<br />
AinF.<br />
Proof: No! For example, Let S R n ,andF be the collection of sets consisting of a<br />
single point x R n . <strong>The</strong>n it is trivially seen that S AF A.<strong>And</strong>ifx is an accumulation<br />
point of S, then x is not an accumulation point of each set A in F.<br />
3.16 Prove that the set S of rational numbers in the inerval 0, 1 cannot be<br />
expressed as the intersection of a countable collection of open sets. Hint: Write<br />
S x 1 , x 2 ,..., assume that S k k1 S k , where each S k is open, and construct a<br />
sequence Q n of closed intervals such that Q n1 Q n S n and such that x n Q n .<br />
<strong>The</strong>n use the Cantor intersection theorem to obtain a contradiction.<br />
Proof: We prove the statement by method of contradiction. Write S x 1 , x 2 ,..., and<br />
assume that S k k1 S k , where each S k is open.<br />
Since x 1 S 1 , there exists a bounded and open interval I 1 S 1 such that x 1 I 1 .<br />
Choose a closed interval Q 1 I 1 such that x 1 Q 1 .SinceQ 1 is an interval, it contains<br />
infinite rationals, call one of these, x 2 .Sincex 2 S 2 , there exists an open interval I 2 S 2<br />
and I 2 Q 1 . Choose a closed interval Q 2 I 2 such that x 2 Q 2 . Suppose Q n has been<br />
constructed so that<br />
1. Q n is a closed interval.<br />
2. Q n Q n1 S n1 .<br />
3. x n Q n .<br />
Since Q n is an interval, it contains infinite rationals, call one of these, x n1 .Since<br />
x n1 S n1 , there exists an open interval I n1 S n1 and I n1 Q n . Choose a closed<br />
interval Q n1 I n1 such that x n1 Q n1 .So,Q n1 satisfies our induction hypothesis, and<br />
the construction can process.<br />
Note that<br />
1. For all n, Q n is not empty.<br />
2. For all n, Q n is bounded since I 1 is bounded.<br />
3. Q n1 Q n .<br />
4. x n Q n .<br />
<strong>The</strong>n n n1 Q n by Cantor Intersection <strong>The</strong>orem.<br />
Since Q n S n , n n1 Q n n n1 S n S. So, we have<br />
S n n1 Q n n n1 Q n <br />
which is absurb since S n n1 Q n by the fact x n Q n . Hence, we have proved that<br />
our assumption does not hold. That is, S the set of rational numbers in the inerval 0, 1<br />
cannot be expressed as the intersection of a countable collection of open sets.<br />
Remark: 1. Often, the property is described by saying Q is not an G set.<br />
2. It should be noted that Q c is an G set.<br />
3. For the famous <strong>The</strong>orem called Cantor Intersection <strong>The</strong>orem, the reader should<br />
see another classical text book, Principles of Mathematical Analysis written by Walter<br />
Rudin, <strong>The</strong>orem 3.10 in page 53.<br />
4. For the method of proof, the reader should see another classical text book, Principles<br />
of Mathematical Analysis written by Walter Rudin, <strong>The</strong>orem 2.43, in page 41.
Covering theorems in R n<br />
3.17 If S R n , prove that the collection of isolated points of S is countable.<br />
Proof: Denote the collection of isolated points of S by F. Letx F, there exists an<br />
n ball Bx, r x x S . Write Q n x 1 , x 2 ,..., then there are many numbers in<br />
Q n lying on Bx, r x x. We choose the smallest index, say m mx, and denote x by<br />
x m .<br />
So, F x m : m P, whereP N, a subset of positive integers. Hence, F is<br />
countable.<br />
3.18 Prove that the set of open disks in the xy plane with center x, x and radius<br />
x 0, x rational, is a countable covering of the set x, y : x 0, y 0.<br />
Proof: Denote the set of open disks in the xy plane with center x, x and radius x 0<br />
by S. Choose any point a, b, wherea 0, and b 0. We want to find an 2 ball<br />
Bx, x, x S which contains a, b. It suffices to find x Q such that<br />
x, x a, b x. Since<br />
x, x a, b x x, x a, b 2 x 2 x 2 2a bx a 2 b 2 0.<br />
Since x 2 2a bx a 2 b 2 x a b 2 2ab, we can choose a suitable rational<br />
number x such that x 2 2a bx a 2 b 2 0sincea 0, and b 0. Hence, for any<br />
point a, b, wherea 0, and b 0, we can find an 2 ball Bx, x, x S which<br />
contains a, b.<br />
That is, S is a countable covering of the set x, y : x 0, y 0.<br />
Remark: <strong>The</strong> reader should give a geometric appearance or draw a graph.<br />
3.19 <strong>The</strong> collection Fof open intervals of the form 1/n,2/n, where n 2, 3, . . . , is an<br />
open covering of the open interval 0, 1. Prove (without using <strong>The</strong>orem 3.31) that no<br />
finite subcollection of F covers 0, 1.<br />
Proof: Write F as 1 ,1, 1 , 2 ,..., 1 2 3 3 n , 2 n ,.... Obviously, F is an open covering<br />
of 0, 1. Assume that there exists a finite subcollection of F covers 0, 1, and thus write<br />
them as F n 1 1<br />
1<br />
, m1 ,...., n 1 1<br />
k<br />
, mk . Choose p 0, 1 so that p min 1ik n 1 i<br />
. <strong>The</strong>n<br />
p n 1 1<br />
i<br />
, mi , where 1 i k. It contracdicts the fact F covers 0, 1.<br />
Remark: <strong>The</strong> reader should be noted that if we use <strong>The</strong>orem 3.31, then we cannot get<br />
the correct proof. In other words, the author T. M. Apostol mistakes the statement.<br />
3.20 Give an example of a set S which is closed but not bounded and exhibit a<br />
coubtable open covering F such that no finite subset of F covers S.<br />
Solution: Let S R 1 , then R 1 is closed but not bounded. <strong>And</strong> let<br />
F n, n 2 : n Z, then F is a countable open covering of S. In additon, it is<br />
trivially seen that no finite subset of F covers S.<br />
3.21 Given a set S in R n with the property that for every x in S there is an n ball Bx<br />
such that Bx S is coubtable. Prove that S is countable.<br />
Proof: Note that F Bx : x S forms an open covering of S. SinceS R n , then<br />
there exists a countable subcover F F of S by Lindelof Covering <strong>The</strong>orem. Write<br />
F Bx n : n N. Since<br />
S S nN Bx n nN S Bx n ,<br />
and<br />
S Bx n is countable by hypothesis.
<strong>The</strong>n S is countable.<br />
Remark: <strong>The</strong> reader should be noted that exercise 3.21 is equivalent to exercise 3.23.<br />
3.22 Prove that a collection of disjoint open sets in R n is necessarily countable. Give an<br />
example of a collection of disjoint closed sets which is not countable.<br />
Proof: Let F be a collection of disjoint open sets in R n , and write Q n x 1 , x 2 ,....<br />
Choose an open set S in F, then there exists an n ball By, r S. In this ball, there<br />
are infinite numbers in Q n . We choose the smallest index, say m my. <strong>The</strong>n we have<br />
F S m : m P N which is countable.<br />
For the example that a collection of disjoint closed sets which is not countable, we give<br />
it as follows. Let G x : x R n , then we complete it.<br />
3.23 Assume that S R n . A point x in R n is said to be condensation point of S if every<br />
n ball Bx has the property that Bx S is not countable. Prove that if S is not<br />
countable, then there exists a point x in S such that x is a condensation point of S.<br />
Proof: It is equivalent to exercise 3.21.<br />
Remark: Compare with two definitions on a condensation point and an accumulation<br />
point, it is easy to know that a condensation point is an accumulation point. However, am<br />
accumulation point is not a condensation point, for example, S 1/n : n N. Wehave<br />
0 is an accumulation point of S, but not a condensation point of S.<br />
3.24 Assume that S R n and assume that S is not countable. Let T denote the set of<br />
condensation points of S. Prove that<br />
(a) S T is countable.<br />
Proof: If S T is uncountable, then there exists a point x in S T such that xisa<br />
condensation point of S T by exercise 3.23. Obviously, x S is also a condensation<br />
point of S. It implies x T. So, we have x S T which is absurb since x S T.<br />
Remark: <strong>The</strong> reader should regard T as a special part of S, and the advantage of T<br />
helps us realize the uncountable set S R n . Compare with Cantor-Bendixon <strong>The</strong>orem<br />
in exercise 3.25.<br />
(b) S T is not countable.<br />
Proof: Suppose S T is countable, then S S T S T is countable by (a)<br />
which is absurb. So, S T is not countable.<br />
(c) T is a closed set.<br />
Proof: Let x be an adherent point of T, then Bx, r T for any r 0. We want to<br />
show x T. That is to show x is a condensation point of S. Claim that Bx, r S is<br />
uncountable for any r 0.<br />
Suppose NOT, then there exists an n ball Bx, d S which is countable. Since x is an<br />
adherent point of T, then Bx, d T . Choose y Bx, d T so that By, Bx, d<br />
and By, S is uncountable. However, we get a contradiction since<br />
By, S is uncountable Bx, d S is countable .<br />
Hence, Bx, r S is uncountable for any r 0. That is, x T. SinceT contains its all<br />
adherent points, T is closed.<br />
(d) T contains no isolated points.<br />
Proof: Let x T, andifx is an isolated point of T, then there exists an n ball Bx, d<br />
such that Bx, d T x. On the other hand, x T means that Bx, d x S is
uncountable. Hence, by exercise 3.23, we know that there exists y Bx, d x S<br />
such that y is a condensation point of Bx, d x S. So,y is a condensation point of<br />
S. It implies y T. It is impossible since<br />
1. y x T.<br />
2. y Bx, d.<br />
3. Bx, d T x.<br />
Hence, x is not an isolated point of T, ifx T. Thatis,T contains no isolatd points.<br />
Remark: Use exercise 3.25, by (c) and (d) we know that T is perfect.<br />
Note that Exercise 3.23 is a special case of (b).<br />
3.25 AsetinR n is called perfect if S S, that is, if S is a closed set which contains<br />
no isolated points. Prove that every uncountable closed set F in R n canbeexpressedinthe<br />
form F A B, where A is perfect and B is countable (Cantor-Bendixon theorem).<br />
Hint. Use Exercise 3.24.<br />
Proof: Let F be a uncountable closed set in R n . <strong>The</strong>n by exercise 3.24,<br />
F F T F T, whereT is the set of condensation points of F. Note that since F is<br />
closed, T F by the fact, a condensation point is an accumulation point. Define<br />
F T A and F T B, then B is countable and A T is perfect.<br />
Remark: 1. <strong>The</strong> reader should see another classical text book, Principles of<br />
Mathematical Analysis written by Walter Rudin, <strong>The</strong>orem 2.43, in page 41. Since the<br />
theorem is famous, we list it below.<br />
<strong>The</strong>orem 2.43 Let P be a nonempty perfect set in R k . <strong>The</strong>n P is uncountable.<br />
<strong>The</strong>orem Modefied 2.43 Let P be a nonempty perfect set in a complete separable<br />
metric space. <strong>The</strong>n P is uncountable.<br />
2. Let S has measure zero in R 1 . Prove that there is a nonempty perfect set P in R 1 such<br />
that P S .<br />
Proof: SinceS has measure zero, there exists a collection of open intervals I k such<br />
that<br />
S I k and |I k | 1.<br />
Consider its complement I k c which is closed with positive measure. Since the<br />
complement has a positive measure, we know that it is uncountable. Hence, by<br />
Cantor-Bendixon <strong>The</strong>orem, we know that<br />
I k c A B, whereA is perfect and B is countable.<br />
So, let A P, wehaveproveit.<br />
Note: From the similar method, we can show that given any set S in R 1 with measure<br />
0 d , there is a non-empty perfect set P such that P S . In particular, S Q,<br />
S the set of algebraic numbers, and so on. In addition, even for cases in R k , it still holds.<br />
Metric Spaces<br />
3.26 In any metric space M, d prove that the empty set and the whole set M are both<br />
open and closed.<br />
proof: In order to show the statement, it suffices to show that M is open and closed<br />
since M M . Letx M, then for any r 0, B M x, r M. Thatis,x is an interior
point of M. Sincx is arbitrary, we know that every point of M is interior. So, M is open.<br />
Let x be an adherent point of M, it is clearly x M since we consider all points lie in<br />
M. Hence, M contains its all adherent points. It implies that M is closed.<br />
Remark: <strong>The</strong> reader should regard the statement as a common sense.<br />
3.27 Consider the following two metrics in R n :<br />
in<br />
d 1 x, y max 1in |x i y i |, d 2 x, y |xi i1<br />
y i |.<br />
In each of the following metric spaces prove that the ball Ba; r has the geometric<br />
appearance indicated:<br />
(a) In R 2 , d 1 , a square with sides parallel to the coordinate axes.<br />
Solution: It suffices to consider the case B0, 0,1. Letx x 1 , x 2 B0, 0,1,<br />
then we have<br />
|x 1 | 1, and |x 2 | 1.<br />
So, it means that the ball B0, 0,1 is a square with sides lying on the coordinate axes.<br />
Hence, we know that Ba; r is a square with sides parallel to the coordinate axes.<br />
(b) In R 2 , d 2 , a square with diagonals parallel to the axes.<br />
Solution: It suffices to consider the case B0, 0,1. Letx x 1 , x 2 B0, 0,1,<br />
then we have<br />
|x 1 x 2 | 1.<br />
So, it means that the ball B0, 0,1 is a square with diagonals lying on the coordinate<br />
axes. Hence, we know that Ba; r is a square with diagonals parallel to the coordinate<br />
axes.<br />
(c) A cube in R 3 , d 1 .<br />
Solution:It suffices to consider the case B0, 0, 0,1. Let<br />
x x 1 , x 2 , x 3 B0, 0, 0,1, then we have<br />
|x 1 | 1, |x 2 | 1, and |x 3 | 1.<br />
So, it means that the ball B0, 0, 0,1 is a cube with length 2. Hence, we know that<br />
Ba; r is a cube with length 2a.<br />
(d) An octahedron in R 3 , d 2 .<br />
Solution: It suffices to consider the case B0, 0, 0,1. Let<br />
x x 1 , x 2 , x 3 B0, 0, 0,1, then we have<br />
|x 1 x 2 x 3 | 1.<br />
It means that the ball B0, 0, 0,1 is an octahedron. Hence, Ba; r is an octahedron.<br />
Remark: <strong>The</strong> exercise tells us one thing that Ba; r may not be an n ball if we<br />
consider some different matrices.<br />
3.28 Let d 1 and d 2 be the metrics of Exercise 3.27 and let x y denote the usual<br />
Euclidean metric. Prove that the following inequalities for all x and y in R n :<br />
d 1 x, y x y d 2 x, y and d 2 x, y n x y nd 1 x, y.<br />
Proof: List the definitions of the three metrics, and compare with them as follows.
<strong>The</strong>n we have<br />
(a)<br />
(b)<br />
(c)<br />
(d)<br />
x y <br />
1. d 1 x, y max 1in |x i y i |.<br />
2. x y in i1<br />
x i y i 2 1/2<br />
.<br />
in<br />
3. d 2 x, y |xi i1<br />
y i |.<br />
d 1 x, y max<br />
1in |x i y i | <br />
<br />
<br />
in<br />
1/2<br />
x i y i 2<br />
i1<br />
in<br />
1/2<br />
x i y i 2<br />
i1<br />
in<br />
|x i y i |<br />
i1<br />
2 1/2<br />
in<br />
1/2<br />
n x y n x i y i 2<br />
i1<br />
d 2 x, y 2 <br />
n n max |x i y i |<br />
1in<br />
d 1 x, y.<br />
in<br />
|x i y i |<br />
i1<br />
2<br />
max |x i y i | 2 1/2<br />
1in<br />
x y.<br />
in<br />
|x i y i | d 2 x, y.<br />
i1<br />
<br />
2 1/2<br />
in<br />
1/2<br />
n x i y i 2<br />
i1<br />
n max<br />
1in |x i y i |<br />
in<br />
x i y i 2 2|x i y i ||x j y j |<br />
i1<br />
1ijn<br />
in<br />
in<br />
x i y i 2 n 1 x i y i 2 by A. P. G. P.<br />
i1<br />
i1<br />
in<br />
n x i y i 2<br />
i1<br />
nx y 2 .<br />
So,<br />
d 2 x, y n x y.<br />
From (a)-(d), we have proved these inequalities.<br />
Remark: 1. Let M be a given set and suppose that M, d and M, d are metric spaces.<br />
We define the metrics d and d are equivalent if, and only if, there exist positive constants<br />
, such that<br />
dx, y dx, y dx, y.<br />
<strong>The</strong> concept is much important for us to consider the same set with different metrics. For
example, in this exercise, Since three metrics are equivalent, it is easy to know that<br />
R k , d 1 , R k , d 2 ,andR k , . are complete. (For definition of complete metric space,<br />
the reader can see this text book, page 74.)<br />
2. It should be noted that on a finite dimensional vector space X, any two norms are<br />
equivalent.<br />
3.29 If M, d is a metric space, define d x, y dx,y . Prove that 1dx,y d isalsoametric<br />
for M. Note that 0 d x, y 1 for all x, yinM.<br />
Proof: In order to show that d isametricforM, we consider the following four steps.<br />
(1) For x M, d x, x 0sincedx, x 0.<br />
(2) For x y, d x, y dx,y 0sincedx, y 0.<br />
1dx,y<br />
(3) For x, y M, d x, y dx,y dy,x d y, x<br />
1dx,y 1dy,x<br />
(4) For x, y, z M,<br />
d dx, y<br />
x, y <br />
1 dx, y 1 1<br />
1 dx, y<br />
1 1 since dx, y dx, z dz, y<br />
1 dx, z dz, y<br />
dx, z dz, y<br />
<br />
1 dx, z dz, y<br />
dx, z<br />
<br />
1 dx, z dz, y dz, y<br />
1 dx, z dz, y<br />
dx, z<br />
<br />
1 dx, z dz, y<br />
1 dz, y<br />
d x, z d z, y<br />
Hence, from (1)-(4), we know that d is also a metric for M. Obviously,<br />
0 d x, y 1 for all x, y in M.<br />
Remark: 1. <strong>The</strong> exercise tells us how to form a new metric from an old metric. Also,<br />
the reader should compare with exercise 3.37. This is another construction.<br />
2. Recall Discrete metric d, we find that given any set nonempty S, S, d is a metric<br />
space, and thus use the exercise, we get another metric space S, d , and so on. Hence,<br />
here is a common sense that given any nonempty set, we can use discrete metric to form<br />
many and many metric spaces.<br />
3.30 Prove that every finite subset of a metric space is closed.<br />
Proof: Let x be an adherent point of a finite subet S x i : i 1, 2, . . . , n of a metric<br />
space M, d. <strong>The</strong>n for any r 0, Bx, r S . Ifx S, then B M x, S where<br />
min 1ijn dx i , x j . It is impossible. Hence, x S. Thatis,S contains its all adherent<br />
points. So, S is closed.<br />
3.31 In a metric space M, d the closed ball of radius r 0 about a point a in M is the<br />
set B a; r x : dx, a r.<br />
(a) Prove that B a; r is a closed set.<br />
Proof: Let x M B a; r, then dx, a r. Consider Bx, , where dx,ar ,<br />
2<br />
then if y Bx, , wehavedy, a dx, a dx, y dx, a dx,ar r. Hence,<br />
2<br />
Bx, M B a; r. That is, every point of M B a; r is interior. So, M B a; r is<br />
open, or equivalently, B a; r is a closed set.
(b) Give an example of a metric space in which B a; r is not the closure of the open<br />
ball Ba; r.<br />
Solution: Consider discrete metric space M, then we have let x M<br />
<strong>The</strong> closure of Ba;1 a<br />
and<br />
B a;1 M.<br />
Hence, if we let a is a proper subset of M, then B a;1 is not the closure of the open ball<br />
Ba;1.<br />
3.32 In a metric space M, if subsets satisfy A S A , where A is the closure of A,<br />
thenAissaidtobedense in S. For example, the set Q of rational numbers is dense in R. If<br />
A is dense in S and if S is dense in T, prove that A is dense in T.<br />
Proof: Since A is dense in S and S is dense in T, wehaveA S and S T. <strong>The</strong>n<br />
A T. Thatis,A is dense in T.<br />
3.33 Refer to exercise 3.32. A metric space M is said to be separable if there is a<br />
countable subset A which is dense in M. For example, R 1 is separable becasue the set Q of<br />
rational numbrs is a countable dense subset. Prove that every Euclidean space R k is<br />
separable.<br />
Proof: Since Q k is a countable subset of R k ,andQ k<br />
R k , then we know that R k is<br />
separable.<br />
3.34 Refer to exercise 3.33. Prove that the Lindelof covering theorem (<strong>The</strong>orem 3.28)<br />
is valid in any separable metric space.<br />
Proof: Let M, d be a separable metric space. <strong>The</strong>n there exists a countable subset<br />
S x n : n N M which is dense in M. Given a set A M, and an open covering F<br />
of A. Write P Bx n , r m : x n S, r m Q.<br />
Claim that if x M, andG is an open set in M which contains x. <strong>The</strong>n<br />
x Bx n , r m G for some Bx n , r m P.<br />
Since x G, there exists Bx, r x G for some r x 0. Note that x clS since S is<br />
dense in M. <strong>The</strong>n, Bx, r x /2 S . So, if we choose x n Bx, r x /2 S and r m Q<br />
with r x /2 r m r x /3, then we have<br />
x Bx n , r m <br />
and<br />
Bx n , r m Bx, r x <br />
since if y Bx n , r m , then<br />
dy, x dy, x n dx n , x<br />
r m r x<br />
2<br />
r x<br />
3 r x<br />
2<br />
r x<br />
So, we have prvoed the claim x Bx n , r m Bx, r x G or some Bx n , r m P.<br />
Use the claim to show the statement as follows. Write A GF G, and let x A, then<br />
there is an open set G in F such that x G. By the claim, there is Bx n , r m : B nm in P<br />
such that x B nm G. <strong>The</strong>re are, of course, infinitely many such B nm corresponding to<br />
each G, but we choose only one of these, for example, the one of smallest index, say<br />
q qx. <strong>The</strong>n we have x B qx G.
<strong>The</strong> set of all B qx obtained as x varies over all elements of A is a countable collection<br />
of open sets which covers A. To get a countable subcollection of F which covers A, we<br />
simply correlate to each set B qx one of the sets G of F which contained B qx . This<br />
complete the proof.<br />
3.35 Refer to exercise 3.32. If A is dense in S and B is open in S, prove that<br />
B clA B, where clA B means the closure of A B.<br />
Hint. Exercise 3.13.<br />
Proof: Since A is dense in S and B is open in S, A S and S B B. <strong>The</strong>n<br />
B S B<br />
A B, B is open in S<br />
clA B<br />
by exercise 3.13.<br />
3.36 Refer to exercise 3.32. If each of A and B is dense in S and if B is open in S, prove<br />
that A BisdenseinS.<br />
Proof: Since<br />
clA B, B is open<br />
clA B by exercise 3.13<br />
S B since A is dense in S<br />
B since B is open in S<br />
then<br />
clA B B<br />
which implies<br />
clA B S<br />
since B is dense in S.<br />
3.37 Given two metric spaces S 1 , d 1 and S 2 , d 2 , ametric for the Cartesian<br />
product S 1 S 2 can be constructed from d 1 d 2 in may ways. For example, if x x 1 , x 2 <br />
and y y 1 , y 2 are in S 1 S 2 , let x, y d 1 x 1 , y 1 d 2 x 2 , y 2 . Prove that is a<br />
metric for S 1 S 2 and construct further examples.<br />
Proof: In order to show that isametricforS 1 S 2 , we consider the following four<br />
steps.<br />
(1) For x x 1 , x 2 S 1 S 2 , x, x d 1 x 1 , x 1 d 2 x 2 , x 2 0 0 0.<br />
(2) For x y, x, y d 1 x 1 , y 1 d 2 x 2 , y 2 0 since if x, y 0, then x 1 y 1<br />
and x 2 y 2 .<br />
(3) For x, y S 1 S 2 ,<br />
x, y d 1 x 1 , y 1 d 2 x 2 , y 2 <br />
d 1 y 1 , x 1 d 2 y 2 , x 2 <br />
y, x.<br />
(4) For x, y, z S 1 S 2 ,<br />
x, y d 1 x 1 , y 1 d 2 x 2 , y 2 <br />
d 1 x 1 , z 1 d 1 z 1 , y 1 d 2 x 2 , z 2 d 2 z 2 , y 2 <br />
d 1 x 1 , z 1 d 2 x 2 , z 2 d 1 z 1 , y 1 d 2 z 2 , y 2 <br />
x, z z, y.
Hence from (1)-(4), we know that isametricforS 1 S 2 .<br />
For other metrics, we define<br />
1 x, y : d 1 x 1 , y 1 d 2 x 2 , y 2 for , 0.<br />
d<br />
2 x, y : d 1 x 1 , y 1 2 x 2 , y 2 <br />
1 d 2 x 2 , y 2 <br />
and so on. (<strong>The</strong> proof is similar with us by above exercises.)<br />
Compact subsets of a metric space<br />
3.38 Assume S T M. <strong>The</strong>n S is compact in M, d if, and only if, S is compact in<br />
the metric subspace T, d.<br />
Proof: Suppose that S is compact in M, d. LetF O : O is open in T be an<br />
open covering of S. SinceO is open in T, there exists the corresponding G which is open<br />
in M such that G T O . It is clear that G forms an open covering of S. Sothereis<br />
a finite subcovering G 1 ,...,G n of S since S is compact in M, d. Thatis,S kn k1 G k .<br />
It implies that<br />
S T S<br />
T kn k1 G k . <br />
kn k1 T G k <br />
kn k1 O k F.<br />
So, we find a fnite subcovering O 1 ,...,O n of S. Thatis,S is compact in T, d.<br />
Suppose that S is compact in T, d. LetG G : G is open in M be an open<br />
covering of S. SinceG T : O is open in T, the collection O forms an open<br />
covering of S. So, there is a finite subcovering O 1 ,...,O n of S since S is compact in<br />
T, d. Thatis,S kn k1 O k . It implis that<br />
S kn k1 O k kn k1 G k .<br />
So, we find a finite subcovering G 1 ,...,G n of S. Thatis,S is compact in M, d.<br />
Remark: <strong>The</strong> exercise tells us one thing that the property of compact is not changed,<br />
but we should note the property of being open may be changed. For example, in the<br />
2 dimensional Euclidean space, an open interval a, b is not open since a, b cannot<br />
contain any 2 ball.<br />
3.39 If S is a closed and T is compact, then S T is compact.<br />
Proof: Since T is compact, T is closed. We have S T is closed. Since S T T, by<br />
<strong>The</strong>orem 3.39, we know that S T is compact.<br />
3.40 <strong>The</strong> intersection of an arbitrary collection of compact subsets of M is compact.<br />
Proof: Let F T : T is compacet in M , and thus consider TF<br />
T, whereF F.<br />
We have TF<br />
T is closed. Choose S F . then we have TF<br />
T S. Hence, by<br />
<strong>The</strong>orem 3.39 TF<br />
T is compact.<br />
3.41 <strong>The</strong> union of a finite number of compact subsets of M is cmpact.<br />
Proof: Denote T k is a compact subset of M : k 1, 2, . . n by S. LetF be an open<br />
covering of kn k1 T k . If there does NOT exist a finite subcovering of kn k1 T k , then there<br />
does not exist a finite subcovering of T m for some T m S. SinceF is also an open<br />
covering of T m , it leads us to get T m is not compact which is absurb. Hence, if F is an open<br />
covering of kn k1 T k , then there exists a finite subcovering of kn k1 T k .So,kn k1 T k is
compact.<br />
3.42 Consider the metric space Q of rational numbers with the Euclidean metric of<br />
R 1 . Let S consists of all rational numbers in the open interval a, b, whereaandbare<br />
irrational. <strong>The</strong>n S is a closed and bounded subset of Q which is not compact.<br />
Proof: Obviously, S is bounded. Let x Q S, then x a, orx b. Ifx a, then<br />
B Q x, d x d, x d Q Q S, whered a x. Similarly, x b. Hence, x is an<br />
interior point of Q S. Thatis,Q S is open, or equivalently, S is closed.<br />
Remark: 1. <strong>The</strong> exercise tells us an counterexample about that in a metric space, a<br />
closed and bounded subset is not necessary to be compact.<br />
2. Here is another counterexample. Let M be an infinite set, and thus consider the<br />
metric space M, d with discrete metric d. <strong>The</strong>nbythefactBx,1/2 x for any x M,<br />
we know that F Bx,1/2 : x M forms an open covering of M. It is clear that there<br />
does not exist a finite subcovering of M. Hence, M is not compact.<br />
3.In any metric space M, d, we have three equivalent conditions on compact which<br />
list them below. Let S M.<br />
(a) Given any open covering of S, there exists a finite subcovering of S.<br />
(b) Every infinite subset of S has an accumulation point in S.<br />
(c) S is totally bounded and complete.<br />
4. It should be note that if we consider the Euclidean spaceR n , d, wehavefour<br />
equivalent conditions on compact which list them below. Let S R n .<br />
Remark (a) Given any open covering of S, there exists a finite subcovering of S.<br />
(b) Every infinite subset of S has an accumulation point in S.<br />
(c) S is totally bounded and complete.<br />
(d) S is bounded and closed.<br />
5. <strong>The</strong> concept of compact is familar with us since it can be regarded as a extension of<br />
Bolzano Weierstrass <strong>The</strong>orem.<br />
Miscellaneous properties of the interior and the boundary<br />
If A and B denote arbitrary subsets of a metric space M, prove that:<br />
3.43 intA M clM A.<br />
Proof: In order to show the statement, it suffices to show that M intA clM A.<br />
1. Let x M intA, we want to show that x clM A, i.e.,<br />
Bx, r M A for all r 0. Suppose Bx, d M A for some d 0. <strong>The</strong>n<br />
Bx, d A which implies that x intA. It leads us to get a conradiction since<br />
x M intA. Hence, if x M intA, then x clM A. Thatis,<br />
M intA clM A.<br />
2. Let x clM A, we want to show that x M intA, i.e., x is not an interior<br />
point of A. Suppose x is an interior point of A, then Bx, d A for some d 0. However,<br />
since x clM A, then Bx, d M A . It leads us to get a conradiction since<br />
Bx, d A. Hence, if x clM A, then x M intA. Thatis,clM A M intA.<br />
From 1 and 2, we know that M intA clM A, or equvilantly,<br />
intA M clM A.
3.44 intM A M A .<br />
Proof: Let B M A, and by exercise 3.33, we know that<br />
M intB clM B<br />
which implies that<br />
intB M clM B<br />
which implies that<br />
intM A M clA.<br />
3.45 intintA intA.<br />
Proof: Since S is open if, and only if, S intS. Hence, Let S intA, wehavethe<br />
equality intintA intA.<br />
3.46<br />
(a) intn<br />
i1 A i n i1 intA i , where each A i M.<br />
Proof: We prove the equality by considering two steps.<br />
(1) Since n<br />
i1<br />
i 1, 2, . . . , n. Hence, intn<br />
i1<br />
A i A i for all i 1,2,...,n, then intn<br />
i1<br />
A i n i1 intA i .<br />
(2) Since intA i A i , then n<br />
i1 intA i n<br />
i1<br />
have<br />
n<br />
i1<br />
From (1) and (2), we know that intn<br />
i1<br />
intA i intn i1 A i .<br />
A i i1<br />
A i .Sincen<br />
i1<br />
n<br />
intA i .<br />
A i intA i for all<br />
intA i is open, we<br />
Remark: Note (2), we use the theorem, a finite intersection of an open sets is open.<br />
Hence, we ask whether an infinite intersection has the same conclusion or not.<br />
Unfortunately, the answer is NO! Just see (b) and (c) in this exercise.<br />
(b) int AF A AF intA, if F is an infinite collection of subsets of M.<br />
Proof: Since AF A A for all A F. <strong>The</strong>n int AF A intA for all A F.<br />
Hence, int AF A AF intA.<br />
(c) Give an example where eqaulity does not hold in (b).<br />
Proof: Let F 1<br />
n , 1 n : n N, then int AF A , and AF intA 0. So,<br />
we can see that in this case, int AF A is a proper subset of AF intA. Hence, the<br />
equality does not hold in (b).<br />
Remark: <strong>The</strong> key to find the counterexample, it is similar to find an example that an<br />
infinite intersection of opens set is not open.<br />
3.47<br />
(a) AF intA int AF A.<br />
Proof: Since intA A, AF intA AF A.Wehave AF intA int AF A<br />
since AF intA is open.<br />
(b) Give an example of a finite collection F in which equality does not hold in (a).<br />
Solution: Consider F Q, Q c , then we have intQ intQ c and<br />
intQ Q c intR 1 R 1 . Hence, intQ intQ c is a proper subset of<br />
intQ Q c R 1 . That is, the equality does not hold in (a).<br />
3.48
(a) intA if A is open or if A is closed in M.<br />
Proof: (1) Suppose that A is open. We prove it by the method of contradiction. Assume<br />
that intA , and thus choose<br />
x intA<br />
intclA clM A<br />
intclA M A<br />
intclA intM A since intS T intS intT.<br />
Since<br />
x intclA Bx, r 1 clA A A <br />
and<br />
x intM A Bx, r 2 M A A c *<br />
we choose r minr 1 , r 2 , then Bx, r A A A c A A c . However,<br />
x A and x A Bx, r A for this r. **<br />
Hence, we get a contradiction since<br />
Bx, r A by (*)<br />
and<br />
Bx, r A by (**).<br />
That is, intA if A is open.<br />
(2) Suppose that A is closed, then we have M A is open. By (1), we have<br />
intM A .<br />
Note that<br />
M A clM A clM M A<br />
clM A clA<br />
A<br />
. Hence, intA if A is closed.<br />
(b) Give an example in which intA M.<br />
Solution: Let M R 1 ,andA Q, then<br />
A clA clM A clQ clQ c R 1 . Hence, we have R 1 intA M.<br />
3.49 If intA intB and if A is closed in M, then intA B .<br />
Proof: Assume that intA B , then choose x intA B, then there exists<br />
Bx, r A B for some r 0. In addition, since intA , we find that Bx, r A.<br />
Hence, Bx, r B A . It implies Bx, r M A . Choose<br />
y Bx, r M A, then we have<br />
y Bx, r By, 1 Bx, r, where 0 1 r<br />
and<br />
y M A By, 2 M A, forsome 2 0.<br />
Choose min 1 , 2 , then we have<br />
By, Bx, r M A<br />
A B A c<br />
B.
That is, intB which is absurb. Hence, we have intA B .<br />
3.50 Give an example in which intA intB but intA B M.<br />
Solution: Consider the Euclidean sapce R 1 , |. |. LetA Q, andB Q c , then<br />
intA intB but intA B R 1 .<br />
3.51 A clA clM A and A M A.<br />
Proof: By the definition of the boundary of a set, it is clear that<br />
A clA clM A. In addition, A clA clM A, and<br />
M A clM A clM M A clM A clA. Hence, we have<br />
A M A.<br />
Remark: It had better regard the exercise as a formula.<br />
3.52 If clA clB , then A B A B.<br />
Proof: We prove it by two steps.<br />
(1) Let x A B, then for all r 0,<br />
Bx, r A B Bx, r A Bx, r B <br />
and<br />
Bx, r A B c Bx, r A c B c *<br />
Note that at least one of Bx, r A and Bx, r B is not empty. Without loss of<br />
generality, we say Bx, r A . <strong>The</strong>n by (*), we have for all r 0,<br />
Bx, r A , andBx, r A c .<br />
That is, x A. Hence, we have proved A B A B.<br />
(2) Let x A B. Without loss of generality, we let x A. <strong>The</strong>n<br />
Bx, r A , andBx, r A c .<br />
Since Bx, r A , wehave<br />
Bx, r A B Bx, r A Bx, r B . **<br />
Claim that Bx, r A B c Bx, r A c B c . Suppsoe NOT, it means that<br />
Bx, r A c B c . <strong>The</strong>n we have<br />
Bx, r A Bx, r clA<br />
and<br />
Bx, r B Bx, r clB.<br />
It implies that by hypothesis, Bx, r clA clB which is absurb. Hence, we have<br />
proved the claim. We have proved that<br />
Bx, r A B by(**).<br />
and<br />
Bx, r A B c .<br />
That is, x A B. Hence, we have proved A B A B.<br />
From (1) and (2), we have proved that A B A B.<br />
Supplement on a separable metric space<br />
Definition (Base) A collection V of open subsets of X is said to be a base for X if the<br />
following is true: For every x X and every open set G X such that x G, we<br />
have<br />
x V G for some .
In other words, every open set in X is the union of a subcollection of V .<br />
<strong>The</strong>orem Every separable metric space has a countable base.<br />
Proof: Let M, d be a separable metric space with S x 1 ,...,x n ,...<br />
satisfying clS M. Consider a collection Bx i , 1 : i, k N, then given any<br />
k<br />
x M and x G, whereG is open in X, wehaveBx, G for some 0.<br />
Since S is dense in M, we know that there is a set Bx i , 1 for some i, k, such that<br />
k<br />
x Bx i , 1 Bx, G. So, we know that M has a countable base.<br />
k<br />
Corollary R k ,wherek N, has a countable base.<br />
Proof: SinceR k is separable, by <strong>The</strong>orem 1, we know that R k has a countable<br />
base.<br />
<strong>The</strong>orem Every compact metric space is separable.<br />
Proof: LetK, d be a compact metric space, and given a radius 1/n, wehave<br />
p<br />
K i1 B x n i ,1/n .<br />
Let S x n i : i, n N , then it is clear S is countable. In order to show that S is<br />
dense in K, givenx K, we want to show that x is an adherent point of S.<br />
Consider Bx, for any 0, there is a point x n i in S such that<br />
B x n i ,1/n Bx, since 1/n 0. Hence, we have shown that Bx, S .<br />
That is, x clS which implies that K clS. So, we finally have K is separable.<br />
Corollary Every compact metric space has a countable base.<br />
Proof: It is immediately from <strong>The</strong>orem 1.<br />
Remark This corallary can be used to show that Arzela-Ascoli <strong>The</strong>orem.
Limits <strong>And</strong> Continuity<br />
Limits of sequence<br />
4.1 Prove each of the following statements about sequences in C.<br />
(a) z n 0if|z| 1; z n diverges if |z| 1.<br />
Proof: For the part: z n 0if|z| 1. Given 0, we want to find that there exists a<br />
positive integer N such that as n N, wehave<br />
|z n 0| .<br />
Note that log|z| 0since|z| 1, hence if we choose a positive integer N log |z|<br />
1,<br />
then as n N, wehave<br />
|z n 0| .<br />
For the part: z n diverges if |z| 1. Assume that z n converges to L, thengiven<br />
1, there exists a positive integer N 1 such that as n N 1 ,wehave<br />
|z n L| 1 <br />
|z| n 1 |L|. *<br />
However, note that log|z| 0since|z| 1, if we choose a positive integer<br />
N max log |z|<br />
1 |L| 1, N 1 , then we have<br />
|z| N 1 |L|<br />
which contradicts (*). Hence, z n diverges if |z| 1.<br />
Remark: 1. Given any complex number z C 0, lim n|z| 1/n 1.<br />
2. Keep lim n n! 1/n in mind.<br />
3. In fact, z n is unbounded if |z| 1. ( z n diverges if |z| 1. ) Since given<br />
M 1, and choose a positive integer N log |z|<br />
M 1, then |z| N M.<br />
(b) If z n 0andifc n is bounded, then c n z n 0.<br />
Proof: Since c n is bounded, say its bound M, i.e., |c n| M for all n N. In<br />
addition, since z n 0, given 0, there exists a positive integer N such that as n N,<br />
we have<br />
|z n 0| /M<br />
which implies that as n N, wehave<br />
|c n z n| M|z n| .<br />
That is, lim n c n z n 0.<br />
(c) z n /n! 0 for every complex z.<br />
Proof: Given a complex z, and thus find a positive integer N such that |z| N/2.<br />
Consider (let n N).<br />
z n<br />
z N<br />
z nN<br />
n! N! N 1N 2 n<br />
Hence, z n /n! 0 for every complex z.<br />
<br />
zN<br />
N!<br />
1<br />
2<br />
nN<br />
0asn .<br />
Remark: <strong>The</strong>re is another proof by using the fact <br />
n1<br />
a n converges which implies<br />
z<br />
a n 0. Since n<br />
n1<br />
converges by ratio test for every complex z, then we have<br />
n!
z n /n! 0 for every complex z.<br />
(d) If a n n 2 2 n, then a n 0.<br />
Proof: Since<br />
0 a n n 2 2 n 2<br />
n 2 2 n 1 n for all n N,<br />
and lim n 1/n 0, we have a n 0asn by Sandwich <strong>The</strong>orem.<br />
4.2 If a n2 a n1 a n /2 for all n 1, show that a n a 1 2a 2 /3. Hint:<br />
a n2 a n1 1 2 a n a n1 .<br />
Proof: Since a n2 a n1 a n /2 for all n 1, we have b n1 b n /2, where<br />
b n a n1 a n . So, we have<br />
n<br />
b n1 1 b1 0asn . *<br />
2<br />
Consider<br />
which implies that<br />
So we have<br />
n1<br />
a n2 a 2 b k 1<br />
2 n<br />
k2<br />
k1<br />
b n <br />
3a n1<br />
2<br />
b k <br />
a 1 2a 2<br />
2<br />
a n a 1 2a 2 /3 by (*).<br />
1<br />
2<br />
a n1 a 1 <br />
.<br />
4.3 If 0 x 1 1andifx n1 1 1 x n for all n 1, prove that x n is a<br />
decreasing sequence with limit 0. Prove also that x n1 /x n 1 2 .<br />
Proof: Claim that 0 x n 1 for all n N. We prove the claim by Mathematical<br />
Induction. As n 1, there is nothing to prove. Suppose that n k holds, i.e., 0 x k 1,<br />
then as n k 1, we have<br />
0 x k1 1 1 x k 1 by induction hypothesis.<br />
So, by Mathematical Induction, we have proved the claim. Use the claim, and then we<br />
have<br />
x<br />
x n1 x n 1 x n 1 x n n x n 1<br />
1 x n 2 1 x n 0since0 x n 1.<br />
So, we know that the sequence x n is a decreasing sequence. Since 0 x n 1 for all<br />
n N, by Completeness of R, (Thatis,a monotonic sequence in R which is bounded is<br />
a convergent sequence.) Hence, we have proved that x n is a convergent sequence,<br />
denoted its limit by x. Note that since<br />
x n1 1 1 x n for all n N,<br />
we have x lim n x n1 lim n 1 1 x n 1 1 x which implies xx 1 0.<br />
Since x n is a decreasing sequence with 0 x n 1 for all n N, we finally have x 0.<br />
For proof of x n1 /x n 1 .Since 2<br />
1 1 x<br />
lim<br />
x0 x 1 2<br />
then we have<br />
*
x n1<br />
x n<br />
1 1 x n<br />
x n<br />
1 2 .<br />
Remark: In (*), it is the derivative of 1 1 x at the point x 0. Of course, we can<br />
prove (*) by L-Hospital Rule.<br />
4.4 Two sequences of positive integers a n and b n are defined recursively by taking<br />
a 1 b 1 1 and equating rational and irrational parts in the equation<br />
a n b n 2 a n1 b n1 2 2 for n 2.<br />
Prove that a n2 2b n2 1 for all n 2. Deduce that a n /b n 2 through values 2 ,and<br />
that 2b n /a n 2 through values 2 .<br />
Proof: Note a n b n 2 a n1 b n1 2 2 for n 2, we have<br />
a n a2 n1 2b2 n1 for n 2, and<br />
b n 2a n1 b n1 for n 2<br />
since if A, B, C, andD N with A B 2 C D 2 , then A C, andB D.<br />
Claim that a n2 2b n2 1 for all n 2. We prove the claim by Mathematical<br />
Induction. As n 2, we have by (*)<br />
a 22 2b 22 a 12 2b 12 2 22a 1 b 1 2 1 2 2 22 2 1. Suppose that as n k 2<br />
holds, i.e., a k2 2b k2 1, then as n k 1, we have by (*)<br />
a2 k1 2b2 k1 a k2 2b k2 2 22a k b k 2<br />
a k4 4b k4 4a k2 b2<br />
k<br />
a k2 2b k2 2<br />
1 by induction hypothesis.<br />
Hence, by Mathematical Induction, we have proved the claim. Note that a n2 2b n2 1for<br />
all n 2, we have<br />
*<br />
2 2<br />
a n 1 2 2<br />
b n bn<br />
and<br />
2<br />
2b n<br />
a n 2 2 2.<br />
a2 n<br />
a<br />
Hence, lim n<br />
1<br />
n b n<br />
2 by lim n b n<br />
0 from (*) through values 2 ,and<br />
2b<br />
lim n<br />
1 n a n<br />
2 by lim n an<br />
0 from (*) through values 2 .<br />
Remark: From (*), we know that a n and b n is increasing since a n N and<br />
b n N. That is, we have lim n a n , and lim n b n .<br />
4.5 A real sequence x n satisfies 7x n1 x n3 6forn 1. If x 1 1 , prove that the<br />
2<br />
sequence increases and find its limit. What happens if x 1 3 or if x 2 1 5 2<br />
Proof: Claim that if x 1 1 ,then0 x 2 n 1 for all n N. We prove the claim by<br />
Mathematical Induction. Asn 1, 0 x 1 1 1. Suppose that n k holds, i.e.,<br />
2<br />
0 x k 1, then as n k 1, we have<br />
0 x k1 x k 3 6<br />
<br />
7<br />
1 7<br />
6 1 by induction hypothesis.<br />
Hence, we have prove the claim by Mathematical Induction. Since<br />
x 3 7x 6 x 3x 1x 2, then
x n1 x n x n 3 6<br />
x<br />
7 n<br />
x n 3 7x n 6<br />
7<br />
0since0 x n 1 for all n N.<br />
It means that the sequence x n (strictly) increasing. Since x n is bounded, by<br />
completeness of R, we know that he sequence x n is convergent, denote its limit by x.<br />
Since<br />
x<br />
x lim n<br />
x n1 lim n3 6<br />
n<br />
<br />
7<br />
x3 6 ,<br />
7<br />
we find that x 3, 1, or 2. Since 0 x n 1 for all n N, we finally have x 1.<br />
Claim that if x 1 3 ,then1 x 2 n 2 for all n N. We prove the claim by<br />
Mathematical Induction. Asn 1, there is nothing to prove. Suppose n k holds, i.e.,<br />
1 x k 2, then as n k 1, we have<br />
1 1 6 x k1 x k 3 6<br />
23 6 2.<br />
7<br />
7 7<br />
Hence, we have prove the claim by Mathematical Induction. Since<br />
x 3 7x 6 x 3x 1x 2, then<br />
x n1 x n x n 3 6<br />
x<br />
7 n<br />
x n 3 7x n 6<br />
7<br />
0since1 x n 2 for all n N.<br />
It means that the sequence x n (strictly) decreasing. Since x n is bounded, by<br />
completeness of R, we know that he sequence x n is convergent, denote its limit by x.<br />
Since<br />
x<br />
x lim n<br />
x n1 lim n3 6<br />
n<br />
<br />
7<br />
x3 6 ,<br />
7<br />
we find that x 3, 1, or 2. Since 1 x n 2 for all n N, we finally have x 1.<br />
Claim that if x 1 5 , then x 2 n 5 for all n N. We prove the claim by<br />
2<br />
Mathematical Induction. Asn 1, there is nothing to prove. Suppose n k holds, i.e.,<br />
x k 5 , then as n k 1,<br />
2<br />
x k1 x k 3 6<br />
5 2 3 6<br />
<br />
7 7<br />
173<br />
56 3 5 2 .<br />
Hence, we have proved the claim by Mathematical Induction. Ifx n was convergent,<br />
say its limit x. <strong>The</strong>n the possibilities for x 3, 1, or 2. However, x n 5 for all n N.<br />
2<br />
So, x n diverges if x 1 5 . 2<br />
Remark: Note that in the case x 1 5/2, we can show that x n is increasing by the<br />
same method. So, it implies that x n is unbounded.<br />
4.6 If |a n| 2and|a n2 a n1 | 1 8 |a 2<br />
n1 a n2 | for all n 1, prove that a n <br />
converges.<br />
Proof: Let a n1 a n b n , then we have |b n1 | 1 8 |b n||a n1 a n| 1 2 |b n|, since<br />
|a n| 2 for all n 1. So, we have |b n1 | 1 2 n |b 1 |. Consider (Let m n)
|a m a n| |a m a m1 a m1 a m2 ...a n1 a n |<br />
|b m1 | ...|b n|<br />
|b 1 | 1<br />
2<br />
m2<br />
... 1<br />
2<br />
n1<br />
, *<br />
then a n is a Cauchy sequence since 1 2 k converges. Hence, we know that a n is a<br />
convergent sequence.<br />
Remark: In this exercise, we use the very important theorem, every Cauchy sequence<br />
in the Euclidean space R k is convergent.<br />
4.7 In a metric space S, d, assume that x n x and y n y. Prove that<br />
dx n , y n dx, y.<br />
Proof: Since x n x and y n y, given 0, there exists a positive integer N such<br />
that as n N, wehave<br />
dx n , x /2 and dy n , y /2.<br />
Hence, as n N, wehave<br />
|dx n , y n dx, y| |dx n , x dy n , y|<br />
dx n , x dy n , y /2 /2<br />
.<br />
So, it means that dx n , y n dx, y.<br />
4.8 Prove that in a compact meric space S, d, every sequence in S has a subsequence<br />
which converges in S. This property also implies that S is compact but you are not required<br />
to prove this. (For a proof see either Reference 4.2 or 4.3.)<br />
Proof: Given a sequence x n S, and let T x 1 , x 2 ,.... If the range of T is finite,<br />
there is nothing to prove. So, we assume that the range of T is infinite. Since S is compact,<br />
and T S, wehaveT has a accumulation point x in S. So, there exists a point y n in T such<br />
that By n , x 1 n . It implies that y n x. Hence, we have proved that every sequence in S<br />
has a subsequence which converges in S.<br />
Remark: If every sequence in S has a subsequence which converges in S, then S is<br />
compact. We give a proof as follows.<br />
Proof: In order to show S is compact, it suffices to show that every infinite subset of S<br />
has an accumulation point in S. Given any infinite subset T of S, and thus we choose<br />
x n T (of course in S). By hypothesis, x n has a subsequence x kn which converges<br />
in S, say its limit x. From definition of limit of a sequence, we know that x is an<br />
accumulation of T. So,S is compact.<br />
4.9 Let A beasubsetofametricspaceS. IfA is complete, prove that A is closed. Prove<br />
that the converse also holds if S is complete.<br />
Proof: Let x be an accumulation point of A, then there exists a sequence x n such that<br />
x n x. Sincex n is convergent, we know that x n is a Cauchy sequence. <strong>And</strong> A is<br />
complete, we have x n converges to a point y A. By uniqueness, we know x y A.<br />
So, A contains its all accumulation points. That is, A is closed.<br />
Suppose that S is complete and A is closed in S. Given any Cauchy sequence<br />
x n A, we want to show x n is converges to a point in A. Trivially, x n is also a<br />
Cauchy sequence in S. SinceS is complete, we know that x n is convergent to a point x in<br />
S. By definition of limit of a sequence, it is easy to know that x is an adherent point of A.<br />
So, x A since A is closed. That is, every Cauchy sequence in A is convergent. So, A is
complete.<br />
Supplement<br />
1. Show that the sequence<br />
lim<br />
n<br />
2n!!<br />
2n 1!! 0.<br />
Proof: Let a and b be positive integers satisfying a b 1. <strong>The</strong>n we have<br />
a!b a!b! a b! ab!. *<br />
So, if we let fn 2n!, then we have, by (*)<br />
2n!!<br />
2n 1!! fn!<br />
fn2n 1! 1<br />
2n 1 0.<br />
Hence, we know that lim n<br />
2. Show that<br />
2n!!<br />
2n1!!<br />
0.<br />
a n 1 1 n 1 2 n 1 n n e 1/2 as n ,<br />
where x means Gauss Symbol.<br />
Proof: Since<br />
x 1 2 x2 log1 x x, for all x 1, 1<br />
we have<br />
k n <br />
<br />
k1<br />
k<br />
n 1 2<br />
k<br />
n<br />
k n <br />
2<br />
log an log 1 n<br />
k<br />
k1<br />
k n <br />
<br />
k1<br />
Consider i 2 n i 1 2 , then by Sandwish <strong>The</strong>orem, we know that<br />
lim n<br />
log a n 1/2<br />
which implies that a n e 1/2 as n .<br />
3. Show that n! 1/n n for all n N. ( n! 1/n as n .)<br />
Proof: We prove it by a special method following Gauss’ method. Consider<br />
n! 1 k n<br />
n n k 1 1<br />
and thus let fk : kn k 1, it is easy to show that fk f1 n for all<br />
k 1, 2, . . . , n. So, we have prove that<br />
n! 2 n n<br />
which implies that<br />
n! 1/n n .<br />
Remark: <strong>The</strong>re are many and many method to show n! 1/n as n . We do not<br />
give a detail proofs about it. But We method it as follows as references.<br />
(a) By A. P. G. P. , we have<br />
k1<br />
n 1<br />
k<br />
n 1<br />
n!<br />
1/n<br />
k<br />
n
and use the fact if a n converges to a, thensois<br />
n<br />
a k k1<br />
n .<br />
(b) Use the fact, by Mathematical Induction, n! 1/n n/3 for all n.<br />
(c) Use the fact, A n /n! 0asn for any real A.<br />
(d) Consider pn n!<br />
n<br />
1/n , and thus taking log pn.<br />
n<br />
(e) Use the famuos formula, a n are positive for all n.<br />
lim inf a n1<br />
a n<br />
lim infa n 1/n lim supa n 1/n lim sup a n1<br />
a n<br />
and let a n n!<br />
n<br />
. n x<br />
(f) <strong>The</strong> radius of the power series k<br />
k0<br />
is .<br />
k!<br />
(g) Ue the fact, 1 1/n n e 1 1/n n1 , then en n e n n! en n1 e n .<br />
(h) More.<br />
Limits of functions<br />
Note. In Exercise 4.10 through 4.28, all functions are real valued.<br />
4.10 Let f be defined on an opne interval a, b and assume x a, b. Consider the<br />
two statements<br />
(a) lim h0 |fx h fx| 0;<br />
(b) lim h0 |fx h fx h| 0.<br />
Prove that (a) always implies (b), and give an example in which (b) holds but (a) does<br />
not.<br />
Proof: (a) Since<br />
lim|fx h fx| 0 lim|fx h fx| 0,<br />
h0 h0<br />
we consider<br />
|fx h fx h|<br />
|fx h fx fx fx h|<br />
|fx h fx| |fx fx h| 0ash 0.<br />
So, we have<br />
lim|fx h fx h| 0.<br />
h0<br />
(b) Let<br />
fx <br />
|x| if x 0,<br />
1ifx 0.<br />
<strong>The</strong>n<br />
lim|f0 h f0 h| 0,<br />
h0<br />
but<br />
lim|f0 h f0| lim||h| 1| 1.<br />
h0 h0<br />
So, (b) holds but (a) does not.<br />
Remark: In case (b), there is another example,
1/|x| if x 0,<br />
gx <br />
0ifx 0.<br />
<strong>The</strong> difference of two examples is that the limit of |g0 h g0| does not exist as h<br />
tends to 0.<br />
4.11 Let f be defined on R 2 .If<br />
lim<br />
x,ya,b<br />
fx, y L<br />
and if the one-dimensional lim xa fx, y and lim yb fx, y both exist, prove that<br />
lim xa<br />
lim<br />
yb<br />
fx, y<br />
lim<br />
yb<br />
lim xa<br />
fx, y L.<br />
Proof: Since lim x,ya,b fx, y L, thengiven 0, there exists a 0 such that as<br />
0 |x, y a, b| , wehave<br />
|fx, y L| /2,<br />
which implies<br />
lim|fx, y L| <br />
yb<br />
which implies<br />
limfx, y L<br />
yb<br />
/2 if 0 |x, y a, b| <br />
lim xa<br />
limfx, y L /2 if 0 |x, y a, b| .<br />
yb<br />
Hence, we have proved lim xa|lim yb x, y L| /2 . Since is arbitrary, we have<br />
which implies that<br />
lim xa<br />
limfx, y L 0<br />
yb<br />
lim xa<br />
limfx, y L 0.<br />
yb<br />
So, lim xa lim yb fx, y L. <strong>The</strong> proof of lim yb lim xa fx, y L is similar.<br />
Remark: 1. <strong>The</strong> exercise is much important since in mathematics, we would encounter<br />
many and many similar questions about the interchange of the order of limits. So, we<br />
should keep the exercise in mind.<br />
2. In the proof, we use the concept: |lim xa fx| 0if,andonlyiflim xa fx 0.<br />
3. <strong>The</strong> hypothesis fx, y L as x, y a, b tells us that every approach form these<br />
points x, y to the point a, b, fx, y approaches to L. Use this concept, and consider the<br />
special approach from points x, y to x, b and thus from x, b to a, b. Note that since<br />
lim yb fx, y exists, it means that we can regrad this special approach as one of approaches<br />
from these points x, y to the point a, b. So, it is natural to have the statement.<br />
4. <strong>The</strong> converse of statement is not necessarily true. For example,<br />
x y if x 0ory 0<br />
fx, y <br />
1 otherwise.<br />
Trivially, we have the limit of fx, y does not exist as x, y 0, 0. However,
lim<br />
y0<br />
fx, y <br />
0ifx 0,<br />
1ifx 0.<br />
and lim<br />
x0<br />
fx, y <br />
0ify 0,<br />
1ify 0.<br />
lim<br />
x0<br />
lim<br />
y0<br />
fx, y<br />
lim<br />
y0<br />
limfx, y 1.<br />
x0<br />
In each of the preceding examples, determine whether the following limits exist and<br />
evaluate those limits that do exist:<br />
lim<br />
x0<br />
limfx, y ; lim<br />
y0 y0<br />
limfx, y ;<br />
x0<br />
lim fx, y.<br />
x,y0,0<br />
Now consider the functions f defined on R 2 as follows:<br />
(a) fx, y x2 y 2<br />
if x, y 0, 0, f0, 0 0.<br />
x 2 y 2<br />
Proof: 1. Since(x 0)<br />
we have<br />
2. Since (y 0)<br />
we have<br />
x<br />
limfx, y lim<br />
2 y 2<br />
x0 x0 x 2 y 2<br />
lim<br />
y0<br />
x<br />
limfx, y lim<br />
2 y 2<br />
y0 y0 x 2 y 2<br />
lim<br />
x0<br />
<br />
limfx, y 1.<br />
x0<br />
<br />
limfx, y 1.<br />
y0<br />
y 2<br />
1 ify 0,<br />
y 2 1ify 0,<br />
x 2<br />
1ifx 0,<br />
x 2 1ifx 0,<br />
3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that<br />
x, y 0, 0 r 0. <strong>The</strong>n<br />
lim fx, y lim x 2 y 2<br />
x,y0,0 x,y0,0 x 2 y 2<br />
lim<br />
r0<br />
r 2 cos 2 sin 2 <br />
r 2<br />
cos 2 sin 2 .<br />
So, if we choose and /2, we find the limit of fx, y does not exist as<br />
x, y 0, 0.<br />
Remark: 1. This case shows that<br />
1 lim<br />
x0<br />
lim<br />
y0<br />
fx, y<br />
lim<br />
y0<br />
lim<br />
x0<br />
fx, y<br />
1<br />
2. Obviously, the limit of fx, y does not exist as x, y 0, 0. Since if it was, then<br />
by (*), (**), and the preceding theorem, we know that<br />
which is absurb.<br />
lim<br />
x0<br />
lim<br />
y0<br />
fx, y<br />
lim<br />
y0<br />
lim<br />
x0<br />
fx, y<br />
**
xy<br />
(b) fx, y <br />
2<br />
if x, y 0, 0, f0, 0 0.<br />
xy 2 xy 2<br />
Proof: 1. Since (x 0)<br />
xy 2<br />
limfx, y lim<br />
0 for all y,<br />
x0 x0<br />
xy 2 x y 2<br />
we have<br />
2. Since (y 0)<br />
we have<br />
lim<br />
y0<br />
limfx, y 0.<br />
x0<br />
xy 2<br />
limfx, y lim<br />
0 for all x,<br />
y0 y0<br />
xy 2 x y 2<br />
lim<br />
x0<br />
limfx, y 0.<br />
y0<br />
3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that<br />
x, y 0, 0 r 0. <strong>The</strong>n<br />
xy 2<br />
fx, y <br />
xy 2 x y 2<br />
So,<br />
r 4 cos 2 sin 2 <br />
r 4 cos 2 sin 2 r 2 2r 2 cossin <br />
cos 2 sin 2 .<br />
cos 2 sin 2 12cossin<br />
r 2<br />
0ifr 0<br />
fx, y<br />
1 if /4 or 5/4.<br />
Hence, we know that the limit of fx, y does not exists as x, y 0, 0.<br />
(c) fx, y 1 x sinxy if x 0, f0, y y.<br />
Proof: 1. Since (x 0)<br />
limfx, y lim 1<br />
x0 x0 x sinxy y *<br />
we have<br />
2. Since (y 0)<br />
we have<br />
limfx, y <br />
y0<br />
lim<br />
y0<br />
lim<br />
x0<br />
limfx, y 0.<br />
x0<br />
1<br />
lim y0 x sinxy 0ifx 0,<br />
lim y0 y 0ifx 0,<br />
limfx, y 0.<br />
y0<br />
3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that<br />
x, y 0, 0 r 0. <strong>The</strong>n
fx, y <br />
1<br />
r cos sinr2 cossin if x r cos 0,<br />
r sin if x r cos 0.<br />
0ifr 0,<br />
<br />
0ifr 0.<br />
So, we know that lim x,y0,0 fx, y 0.<br />
Remark: In (*) and (**), we use the famuos limit, that is,<br />
lim sin x<br />
x0 x 1.<br />
<strong>The</strong>re are some similar limits, we write them without proofs.<br />
(a) lim t t sin1/t 1.<br />
(b) lim x0 x sin1/x 0.<br />
(c) lim x0 ,ifb 0.<br />
(d) fx, y <br />
sinax<br />
sinbx<br />
a b<br />
x y sin1/x sin1/y if x 0andy 0,<br />
0 if x 0ory 0.<br />
**<br />
Proof: 1. Since (x 0)<br />
x y sin1/x sin1/y x sin1/x sin1/y y sin1/x sin1/y if y 0<br />
fx, y <br />
0ify 0<br />
we have if y 0, the limit fx, y does not exist as x 0, and if y 0, lim x0 fx, y 0.<br />
Hence, we have (x 0, y 0)<br />
2. Since (y 0)<br />
lim<br />
y0<br />
lim<br />
x0<br />
fx, y<br />
does not exist.<br />
x y sin1/x sin1/y x sin1/x sin1/y y sin1/x sin1/y if x 0<br />
fx, y <br />
0ifx 0<br />
we have if x 0, the limit fx, y does not exist as y 0, and if x 0, lim y0 fx, y 0.<br />
Hence, we have (x 0, y 0)<br />
3. (x, y 0, 0) Consider<br />
we have<br />
lim<br />
x0<br />
|fx, y| <br />
lim<br />
y0<br />
fx, y<br />
does not exist.<br />
|x y| if x 0andy 0,<br />
0ifx 0ory 0.<br />
lim fx, y 0.<br />
x,y0,0<br />
sinxsiny<br />
tan xtan y<br />
,iftanx tan y,<br />
(e) fx, y <br />
cos 3 x if tan x tan y.<br />
Proof: Since we consider the three approaches whose tend to 0, 0, we may assume<br />
that x, y /2, /2. and note that in this assumption, x y tan x tan y. Consider<br />
1. (x 0)
So,<br />
So,<br />
2. (y 0)<br />
lim<br />
x0<br />
fx, y <br />
lim<br />
y0<br />
fx, y <br />
sinxsiny<br />
lim x0 tan xtan y<br />
cosy if x y.<br />
1 if x y.<br />
lim<br />
y0<br />
limfx, y 1.<br />
x0<br />
sinxsiny<br />
lim y0 tan xtan y<br />
cosx if x y.<br />
cos 3 x if x y.<br />
lim<br />
x0<br />
limfx, y 1.<br />
y0<br />
3. Let x r cos and y r sin , where 0 2, and note that<br />
x, y 0, 0 r 0. <strong>The</strong>n<br />
lim<br />
lim fx, y <br />
x,y0,0<br />
sinr cos sinr sin<br />
r0 if cos sin .<br />
tanr cos tanr sin<br />
lim r0 cos 3 r cos if cos sin .<br />
1ifcos sin , byL-Hospital Rule.<br />
<br />
1ifcos sin .<br />
So, we know that lim x,y0,0 fx, y 1.<br />
and<br />
then<br />
So,<br />
Remark: 1. <strong>The</strong>re is another proof about (e)-(3). Consider<br />
x y<br />
sin x sin y 2cos sin<br />
x y<br />
2 2<br />
lim<br />
lim fx, y <br />
x,y0,0<br />
That is, lim x,y0,0 fx, y 1.<br />
tan x tan y sin x cosx sin y<br />
cosy ,<br />
sin x sin y<br />
tan x tan y<br />
xy<br />
cos<br />
2<br />
cosxcosy<br />
cos xy .<br />
2<br />
x,y0,0<br />
cos<br />
xy<br />
cos xcos y<br />
2<br />
cos<br />
xy<br />
2<br />
lim x,y0,0 cos 3 x 1ifx y.<br />
1ifx y,<br />
2. In the process of proof, we use the concept that we write it as follows. Since its proof<br />
is easy, we omit it. If<br />
or<br />
L<br />
lim fx, y <br />
x,ya,b<br />
if x y<br />
L if x y
then we have<br />
lim fx, y L if x 0andy 0,<br />
x,ya,b L if x 0ory 0.<br />
lim fx, y L.<br />
x,ya,b<br />
4.12 If x 0, 1 prove that the following limit exists,<br />
lim<br />
m<br />
lim n cos2n m!x ,<br />
and that its value is 0 or 1, according to whether x is irrational or rational.<br />
Hence,<br />
Proof: If x is rational, say x q/p, whereg. c. d. q, p 1, then p!x N. So,<br />
lim n<br />
cos 2n m!x <br />
lim<br />
m<br />
1ifm p,<br />
0ifm p.<br />
lim n cos2n m!x 1.<br />
If x is irrational, then m!x N for all m N. So, cos 2n m!x 1 for all irrational x.<br />
Hence,<br />
lim n<br />
cos 2n m!x 0 lim m<br />
Continuity of real-valued functions<br />
lim n<br />
cos 2n m!x 0.<br />
4.13 Let f be continuous on a, b and let fx 0 when x is rational. Prove that<br />
fx 0 for every x a, b.<br />
Proof: Given any irrational number x in a, b, and thus choose a sequence x n Q<br />
such that x n x as n . Note that fx n 0 for all n. Hence,<br />
0 lim n<br />
0<br />
lim n<br />
fx n <br />
f lim n<br />
x n by continuity of f at x<br />
fx.<br />
Since x is arbitrary, we have shown fx 0 for all x a, b. Thatis,f is constant 0.<br />
Remark: Here is another good exercise, we write it as a reference. Let f be continuous<br />
on R, andiffx fx 2 , then f is constant.<br />
Proof: Since fx f x 2 fx 2 fx, we know that f is an even function. So,<br />
in order to show f is constant on R, it suffices to show that f is constant on 0, . Given<br />
any x 0, , sincefx 2 fx for all x R, wehavefx 1/2n fx for all n. Hence,<br />
fx lim n<br />
fx<br />
lim n<br />
fx 1/2n <br />
f lim n<br />
x 1/2n by continuity of f at 1<br />
f1 since x 0.<br />
So, we have fx f1 : c for all x 0, . In addition, given a sequence<br />
x n 0, such that x n 0, then we have
c lim n<br />
c<br />
lim n<br />
fx n <br />
f lim n<br />
x n by continuity of f at 0<br />
f0<br />
From the preceding, we have proved that f is constant.<br />
4.14 Let f be continuous at the point a a 1 , a 2 ,...,a n R n . Keep a 2 , a 3 ,...,a n fixed<br />
and define a new function g of one real variable by the equation<br />
gx fx, a 2 ,...,a n .<br />
Prove that g is continuous at the point x a 1 . (This is sometimes stated as follows: A<br />
continuous function of n variables is continuous in each variable separately.)<br />
Proof: Given 0, there exists a 0 such that as y Ba; D, whereD is a<br />
domain of f, wehave<br />
|fy fa| . *<br />
So, as |x a 1 | , which implies |x, a 2 ,...,a n a 1 , a 2 ,...,a n | , wehave<br />
|gx ga 1 | |fx, a 2 ,...,a n fa 1 , a 2 ,...,a n | .<br />
Hence, we have proved g is continuous at x a 1<br />
Remark: Here is an important example like the exercise, we write it as follows. Let<br />
j : R n R n ,and j : x 1 , x 2 ,...,x n 0,.,x j ,..,0. <strong>The</strong>n j is continuous on R n for<br />
1 j n. Note that j is called a projection. Note that a projection P is sometimes<br />
defined as P 2 P.<br />
Proof: Given any point a R n ,and given 0, and choose , then as<br />
x Ba; , wehave<br />
| j x j a| |x j a j | x a for each 1 j n<br />
Hence, we prove that j x is continuous on R n for 1 j n.<br />
4.15 Show by an example that the converse of statement in Exercise 4.14 is not true in<br />
general.<br />
Proof: Let<br />
fx, y <br />
x y if x 0ory 0<br />
1otherwise.<br />
Define g 1 x fx,0 and g 2 y f0, y, then we have<br />
limg 1 x 0 g 1 0<br />
x0<br />
and<br />
limg 2 y 0 g 2 0.<br />
y0<br />
So, g 1 x and g 2 y are continuous at 0. However, f is not continuous at 0, 0 since<br />
limfx, x 1 0 f0, 0.<br />
x0<br />
Remark: 1. For continuity, if f is continuous at x a, then it is NOT necessary for us<br />
to have<br />
lim xa<br />
fx fa
this is because a can be an isolated point. However, if a is an accumulation point, we then<br />
have<br />
f is continuous at a if, and only if, lim xa<br />
fx fa.<br />
4.16 Let f, g, andh be defined on 0, 1 as follows:<br />
fx gx hx 0, whenever x is irrational;<br />
fx 1andgx x, whenever x is rational;<br />
hx 1/n, ifx is the rational number m/n (in lowest terms);<br />
h0 1.<br />
Prove that f is not continuous anywhere in 0, 1, that g is continuous only at x 0, and<br />
that h is continuous only at the irrational points in 0, 1.<br />
Proof: 1. Write<br />
0ifx R Q 0, 1,<br />
fx <br />
1ifx Q 0, 1.<br />
Given any x R Q 0, 1, andy Q 0, 1, and thus choose x n Q 0, 1<br />
such that x n x, andy n R Q 0, 1 such that y n y. Iff is continuous at x,<br />
then<br />
1 lim n<br />
fx n <br />
f lim n<br />
x n by continuity of f at x<br />
fx<br />
0<br />
which is absurb. <strong>And</strong> if f is continuous at y, then<br />
0 lim n<br />
fy n <br />
f lim n<br />
y n by continuity of f at y<br />
fy<br />
1<br />
which is absurb. Hence, f is not continuous on 0, 1.<br />
2. Write<br />
0ifx R Q 0, 1,<br />
gx <br />
x if x Q 0, 1.<br />
Given any x R Q 0, 1, and choose x n Q 0, 1 such that x n x. <strong>The</strong>n<br />
x<br />
lim n<br />
x n<br />
lim n<br />
gx n <br />
limg lim n<br />
x n by continuity of g at x<br />
gx<br />
0<br />
which is absurb since x is irrational. So, f is not continous on R Q 0, 1.<br />
Given any x Q 0, 1, and choose x n R Q 0, 1 such that x n x. Ifg is
continuous at x, then<br />
0<br />
lim n<br />
gx n <br />
g lim n<br />
x n by continuity of f at x<br />
gx<br />
x.<br />
So, the function g may be continuous at 0. In fact, g is continuous at 0 which prove as<br />
follows. Given 0, choose , as|x| , wehave<br />
|gx g0| |gx| |x| . So,g is continuous at 0. Hence, from the preceding,<br />
we know that g is continuous only at x 0.<br />
3. Write<br />
1ifx 0,<br />
hx 0ifx R Q 0, 1,<br />
1/n if x m/n, g. c. d. m, n 1.<br />
Consider a 0, 1 and given 0, there exists the largest positive integer N such that<br />
N 1/. LetT x : hx , then<br />
0, 1 x : hx 1 x : hx 1/2...x : hx 1/N if 1,<br />
T <br />
if 1.<br />
Note that T is at most a finite set, and then we can choose a 0 such that<br />
a , a acontains no points of T and a , a 0, 1. So,if<br />
x a , a a, wehavehx . It menas that<br />
lim xa<br />
hx 0.<br />
Hence, we know that h is continuous at x 0, 1 R Q. For two points x 1, and<br />
y 0, it is clear that h is not continuous at x 1, and not continuous at y 1bythe<br />
method mentioned in the exercise of part 1 and part 2. Hence, we have proved that h is<br />
continuous only at the irrational points in 0, 1.<br />
Remark: 1. Sometimes we call f Dirichlet function.<br />
2. Here is another proof about g, we write it down to make the reader get more.<br />
Proof: Write<br />
0ifx R Q 0, 1,<br />
gx <br />
x if x Q 0, 1.<br />
Given a 0, 1, andifg is continuous at a, then given 0 a, there exists a 0<br />
such that as x a , a 0, 1, wehave<br />
|gx ga| .<br />
If a R Q, choose 0 so that a Q. <strong>The</strong>n a a , a which<br />
implies |ga ga| |ga | a a. But it is impossible.<br />
If a Q, choose 0 so that a R Q. a a , a which<br />
implies |ga ga| |a| a a. But it is impossible.<br />
If a 0, given 0 and choose , thenas0 x , wehave<br />
|gx g0| |gx| |x| x . It means that g is continuous at 0.<br />
4.17 For each x 0, 1, letfx x if x is rational, and let fx 1 x if x is
irrational. Prove that:<br />
(a) ffx x for all x in 0, 1.<br />
Proof: If x is rational, then ffx fx x. <strong>And</strong>ifx is irrarional, so is<br />
1 x 0, 1. <strong>The</strong>n ffx f1 x 1 1 x x. Hence, ffx x for all x in<br />
0, 1.<br />
(b) fx f1 x 1 for all x in 0, 1.<br />
Proof: If x is rational, so is 1 x 0, 1. <strong>The</strong>n fx f1 x x 1 x 1. <strong>And</strong><br />
if x is irrarional, so is 1 x 0, 1. <strong>The</strong>n fx f1 x 1 x 1 1 x 1.<br />
Hence, fx f1 x 1 for all x in 0, 1.<br />
(c) f is continuous only at the point x 1 . 2<br />
Proof: If f is continuous at x, then choose x n Q and y n Q c such that x n x,<br />
and y n x. <strong>The</strong>n we have, by continuity of f at x,<br />
fx f lim n<br />
x n<br />
lim n<br />
fx n lim n<br />
x n x<br />
and<br />
fx f lim n<br />
y n lim n<br />
fy n lim n<br />
1 y n 1 x.<br />
So, x 1/2 is the only possibility for f. Given 0, we want to find a 0 such that as<br />
x 1/2 ,1/2 0, 1, wehave<br />
|fx f1/2| |fx 1/2| .<br />
Choose 0 so that 1/2 ,1/2 0, 1, then as<br />
x 1/2 ,1/2 0, 1, wehave<br />
|fx 1/2| |x 1/2| if x Q,<br />
|fx 1/2| |1 x 1/2| |1/2 x| if x Q c .<br />
Hence, we have proved that f is continuous at x 1/2.<br />
(d) f assumes every value between 0 and 1.<br />
Proof: Given a 0, 1, wewanttofindx 0, 1 such that fx a. Ifa Q, then<br />
choose x a, wehavefx a a. Ifa R Q, then choose x 1 a R Q, we<br />
have fx 1 a 1 1 a a.<br />
Remark: <strong>The</strong> range of f on 0, 1 is 0, 1. In addition, f is an one-to-one mapping<br />
since if fx fy, then x y. (<strong>The</strong> proof is easy, just by definition of 1-1, so we omit it.)<br />
(e) fx y fx fy is rational for all x and y in 0, 1.<br />
Proof: We prove it by four steps.<br />
1. If x Q and y Q, then x y Q. So,<br />
fx y fx fy x y x y 0 Q.<br />
2. If x Q and y Q c , then x y Q c .So,<br />
fx y fx fy 1 x y x 1 y 2x Q.<br />
3. If x Q c and y Q, then x y Q c .So,<br />
fx y fx fy 1 x y 1 x y 2y Q.<br />
4. If x Q c and y Q c , then x y Q c or x y Q. So,
fx y fx fy <br />
1 x y 1 x 1 y 1 Q if x y Q c ,<br />
x y 1 x 1 y 2 Q if x y Q.<br />
Remark: Here is an interesting question about functions. Let f : R 0, 1 R. Iff<br />
satisfies that<br />
fx f x 1<br />
x 1 x,<br />
then fx x3 x 2 1<br />
. 2xx1<br />
Proof: Let x x1<br />
x , then we have 2 x 1<br />
x1 ,and3 x x. So,<br />
fx f x 1<br />
x fx fx 1 x *<br />
which implies that<br />
fx f 2 x 1 x **<br />
and<br />
f 2 x f 3 x f 2 x fx 1 2 x. ***<br />
So, by (*), (**), and (***), we finally have<br />
fx 1 2 1 x x 2 x<br />
x3 x 2 1<br />
2xx 1 .<br />
4.18 Let f be defined on R and assume that there exists at least one x 0 in R at which f<br />
is continuous. Suppose also that, for every x and y in R, f satisfies the equation<br />
fx y fx fy.<br />
Prove that there exists a constant a such that fx ax for all x.<br />
Proof: Let f be defined on R and assume that there exists at least one x 0 in R at which f<br />
is continuous. Suppose also that, for every x and y in R, f satisfies the equation<br />
fx y fx fy.<br />
Prove that there exists a constant a such that fx ax for all x.<br />
Proof: Suppose that f is continuous at x 0 , and given any r R. Since<br />
fx y fx fy for all x, then<br />
fx fy x 0 fr, wherey x r x 0 .<br />
Note that y x 0 x r, then<br />
lim xr<br />
fx lim xr<br />
fy x 0 fr<br />
yx0<br />
lim fy x 0 fr<br />
fr since f is continuous at x 0 .<br />
So, f is continuous at r. Sincer is arbitrary, we have f is continuous on R. Definef1 a,<br />
andthensincefx y fx fy, wehave<br />
f1 f 1 m .. m 1 mtimes<br />
mf<br />
1 m<br />
f 1 m f1<br />
m *
In addition, since f1 f1 by f0 0, we have<br />
f1 f 1 m .. m<br />
1<br />
mf<br />
1 m<br />
f 1 m<br />
mtimes<br />
f1<br />
m *’<br />
Thus we have<br />
f m n f1/m ...1/m ntimes nf1/m m n f1 by (*) and (*’) **<br />
So, given any x R, and thus choose a sequence x n Q with x n x. <strong>The</strong>n<br />
fx f lim n<br />
x n<br />
lim n<br />
fx n by continuity of f on R<br />
lim n<br />
x n f1 by (**)<br />
xf1<br />
ax.<br />
Remark: <strong>The</strong>re is a similar statement. Suppose that fx y fxfy for all real x<br />
and y.<br />
(1) If f is differentiable and non-zero, prove that fx e cx ,wherec is a constant.<br />
Proof: Note that f0 1sincefx y fxfy and f is non-zero. Since f is<br />
differentiable, we define f 0 c. Consider<br />
fx h fx fh f0<br />
fx fxf<br />
h<br />
h<br />
0 cfx as h 0,<br />
we have for every x R, f x cfx. Hence,<br />
fx Ae cx .<br />
Since f0 1, we have A 1. Hence, fx e cx ,wherec is a constant.<br />
Note: (i) If for every x R, f x cfx, then fx Ae cx .<br />
Proof: Since f x cfx for every x, we have for every x,<br />
f x cfxe cx 0 e cx fx 0.<br />
We note that by Elementary Calculus, e cx fx is a constant function A. So,fx Ae cx<br />
for all real x.<br />
(ii) Suppose that fx y fxfy for all real x and y. Iffx 0 0forsomex 0 , then<br />
fx 0 for all x.<br />
Proof: Suppose NOT, then fa 0forsomea. However,<br />
0 fx 0 fx 0 a a fx 0 afa 0.<br />
Hence, fx 0 for all x.<br />
(iii) Suppose that fx y fxfy for all real x and y. Iff is differentiable at x 0 for<br />
some x 0 , then f is differentiable for all x. <strong>And</strong> thus, fx C R.<br />
Proof: Since
fx h fx<br />
h<br />
fx 0 h x x 0 fx 0 x x 0 <br />
h<br />
fx x 0 fx 0 h fx 0 <br />
fx x 0 f x 0 as h 0,<br />
h<br />
we have f x is differentiable and f x fx x 0 f x 0 for all x. <strong>And</strong> thus we have<br />
fx C R.<br />
(iv) Here is another proof by (iii) and Taylor <strong>The</strong>orem with Remainder term R n x.<br />
Proof: Since f is differentiable, by (iii), we have f n x f 0 n fx for all x.<br />
Consider x r, r, then by Taylor <strong>The</strong>orem with Remainder term R n x,<br />
n<br />
fx <br />
k0<br />
<strong>The</strong>n<br />
f k 0<br />
k!<br />
x k R n x, whereR n x : fn1 <br />
n 1! xn1 , 0, x or x,0,<br />
|R n x| fn1 <br />
n 1! xn1<br />
<br />
f 0 n1 f<br />
n 1!<br />
f<br />
<br />
0r n1<br />
n 1!<br />
0asn .<br />
Hence, we have for every x r, r<br />
<br />
fx <br />
k0<br />
f0<br />
x n1<br />
M, whereM max<br />
xr,r |fx|<br />
f<br />
k<br />
0<br />
k!<br />
<br />
<br />
k0<br />
x k<br />
f 0x k<br />
k!<br />
e cx ,wherec : f 0.<br />
Since r is arbitrary, we have proved that fx e cx for all x.<br />
(2) If f is continuous and non-zero, prove that fx e cx ,wherec is a constant.<br />
Proof: Since fx y fxfy, wehave<br />
0 f1 f 1 n ... 1 n<br />
n f 1<br />
ntimes n f 1 n f1 1/n *<br />
and (note that f1 f1 1 by f0 1, )<br />
0 f1 f 1 n ... 1 n f n<br />
1<br />
ntimes n f 1 n f1 1/n *’<br />
f m n f 1 n ... 1 n mtimes<br />
f 1 n<br />
m<br />
f1<br />
m n<br />
by (*) and (*’) **<br />
So, given any x R, and thus choose a sequence x n Q with x n x. <strong>The</strong>n
lim<br />
r n0<br />
fx f lim n<br />
x n<br />
lim n<br />
fx n by continuity of f<br />
lim n<br />
f1 xn by (**)<br />
f1 x<br />
e cx , where log f1 c.<br />
Note: (i) We can prove (2) by the exercise as follows. Note that fx 0 for all x by<br />
the remark (1)-(ii) Consider the composite function gx log fx, then<br />
gx y log fx y log fxfy log fx log fy gx gy. Since log and f<br />
are continuous on R, its composite function g is continuous on R. Use the exercise, we<br />
have gx cx for some c. <strong>The</strong>refore, fx e gx e cx .<br />
(ii) We can prove (2) by the remark (1) as follows. It suffices to show that this f is<br />
differentiable at 0 by remark (1) and (1)-(iii). Sincef m n f1 m n<br />
then for every real r,<br />
fr f1 r a<br />
by continuity of f. Note that lim r b r<br />
r0 r exists. Given any sequence r n <br />
with r n 0, and thus consider<br />
fr n f0<br />
1<br />
1<br />
r n<br />
f1rn r n<br />
f1rn rn<br />
r n<br />
exists,<br />
we have f is differentiable at x 0. So,byremark(1),wehavefx e cx .<br />
(3) Give an example such that f is not continuous on R.<br />
Solution: Consider gx y gx gy for all x, y. <strong>The</strong>n we have gq qg1,<br />
where q Q. ByZorn’s Lemma, we know that every vector space has a basis<br />
v : I. Note that v : I is an uncountable set, so there exists a convergent<br />
sequence s n v : I. Hence, S : v : I s n <br />
n1<br />
sn<br />
n <br />
n1<br />
is a new<br />
basis of R over Q. Givenx, y R, and we can find the same N such that<br />
N<br />
x <br />
k1<br />
N<br />
q k v k and y p k v k ,wherev k S<br />
k1<br />
Define the sume<br />
N<br />
x y : p k q k v k<br />
k1<br />
By uniqueness, we define gx to be the sum of coefficients, i.e.,<br />
N<br />
gx : q k .<br />
k1<br />
Note that<br />
g<br />
s n<br />
1 for all n lim n<br />
g<br />
and<br />
s n 0asn <br />
Hence, g is not continuous at x 0 since if it was, then<br />
s n 1
1 lim n<br />
g<br />
s n<br />
g lim<br />
s n n by continuity of g at 0<br />
g0<br />
0<br />
which is absurb. Hence, g is not continuous on R by the exercise. To find such f, it suffices<br />
to consider fx e gx .<br />
Note: Such g (or f) isnot measurable by Lusin <strong>The</strong>orem.<br />
4.19 Let f be continuous on a, b and define g as follows: ga fa and, for<br />
a x b, letgx be the maximum value of f in the subinterval a, x. Show that g is<br />
continuous on a, b.<br />
Proof: Define gx maxft : t a, x, and choose any point c a, b, wewant<br />
to show that g is continuous at c. Given 0, we want to find a 0 such that as<br />
x c , c a, b, wehave<br />
|gx gc| .<br />
Since f is continuous at x c, then there exists a 0 such that as<br />
x c , c a, b, wehave<br />
fc /2 fx fc /2. *<br />
Consider two cases as follows.<br />
(1) maxft : t a, c a, b fp 1 ,wherep 1 c .<br />
As x c , c a, b, wehavegx fp 1 and gc fp 1 .<br />
Hence, |gx gc| 0.<br />
(2) maxft : t a, c a, b fp 1 ,wherep 1 c .<br />
As x c , c a, b, wehaveby(*)fc /2 gx fc /2.<br />
Hence, |gx gc| .<br />
So, if we choose ,thenforx c , c a, b,<br />
|gx gc| by (1) and (2).<br />
Hence, gx is continuous at c. <strong>And</strong> since c is arbitrary, we have gx is continuous on<br />
a, b.<br />
Remark: It is the same result for minft : t a, x by the preceding method.<br />
4.20 Let f 1 ,...,f m be m real-valued functions defined on R n . Assume that each f k is<br />
continuous at the point a of S. Define a new function f as follows: For each x in S, fx is<br />
the largest of the m numbers f 1 x,...,f m x. Discuss the continuity of f at a.<br />
Proof: Assume that each f k is continuous at the point a of S, then we have f i f j and<br />
|f i f j | are continuous at a, where 1 i, j m. Sincemaxa, b ab|ab| , then<br />
2<br />
maxf 1 , f 2 is continuous at a since both f 1 f 2 and |f 1 f 2 | are continuous at a. Define<br />
fx maxf 1 ,...f m ,useMathematical Induction to show that fx is continuous at<br />
x a as follows. As m 2, we have proved it. Suppose m k holds, i.e., maxf 1 ,...f k is<br />
continuous at x a. <strong>The</strong>n as m k 1, we have<br />
maxf 1 ,...f k1 maxmaxf 1 ,...f k , f k1 <br />
is continuous at x a by induction hypothesis. Hence, by Mathematical Induction, we<br />
have prove that f is continuos at x a.<br />
It is possible that f and g is not continuous on R whihc implies that maxf, g is<br />
continuous on R. For example, let fx 0ifx Q, andfx 1ifx Q c and gx 1
if x Q, andgx 0ifx Q c .<br />
Remark: It is the same rusult for minf 1 ,...f m since maxa, b mina, b a b.<br />
4.21 Let f : S R be continuous on an open set in R n , assume that p S, and assume<br />
that fp 0. Prove that there is an n ball Bp; r such that fx 0 for every x in the<br />
ball.<br />
Proof: Since p S is an open set in R n , there exists a 1 0 such that Bp, 1 S.<br />
Since fp 0, given fp 0, then there exists an n ball Bp; <br />
2 2 such that as<br />
x Bp; 2 S, wehave<br />
fp<br />
fp fx fp 3fp .<br />
2<br />
2<br />
Let min 1 , 2 , then as x Bp; , wehave<br />
fx fp 0.<br />
2<br />
Remark: <strong>The</strong> exercise tells us that under the assumption of continuity at p, we roughly<br />
have the same sign in a neighborhood of p, iffp 0 or fp 0.<br />
4.22 Let f be defined and continuous on a closed set S in R. Let<br />
A x : x S and fx 0 .<br />
Prove that A is a closed subset of R.<br />
Proof: Since A f 1 0, andf is continous on S, wehaveA is closed in S. <strong>And</strong><br />
since S is closed in R, we finally have A is closed in R.<br />
Remark: 1. Roughly speaking, the property of being closed has Transitivity. Thatis,<br />
in M, d let S T M, ifS is closed in T, andT is closed in M, then S is closed in M.<br />
Proof: Let x be an adherent point of S in M, then B M x, r S for every r 0.<br />
Hence, B M x, r T for every r 0. It means that x is also an adherent point of T in<br />
M. SinceT is closed in M, we find that x T. Note that since B M x, r S for every<br />
r 0, we have (S T)<br />
B T x, r S B M x, r T S B M x, r S T B M x, r S .<br />
So, we have x is an adherent point of S in T. <strong>And</strong> since S is closed in T, wehavex S.<br />
Hence, we have proved that if x is an adherent point of S in M, then x S. Thatis,S is<br />
closed in M.<br />
Note: (1) Another proof of remark 1, since S is closed in T, there exists a closed subset<br />
U in Msuch that S U T, and since T is closed in M, wehaveS is closed in M.<br />
(2) <strong>The</strong>re is a similar result, in M, d let S T M, ifS is open in T, andT is open<br />
in M, then S is open in M. (Leave to the reader.)<br />
2. Here is another statement like the exercise, but we should be cautioned. We write it<br />
as follows. Let f and g be continuous on S, d 1 into T, d 2 .LetA x : fx gx,<br />
show that A is closed in S.<br />
Proof: Let x be an accumulation point of A, then there exists a sequence x n A<br />
such that x n x. So, we have fx n gx n for all n. Hence, by continuity of f and g, we<br />
have<br />
fx f lim n<br />
x n lim n<br />
fx n lim n<br />
gx n g lim n<br />
x n gx.<br />
Hence, x A. Thatis,A contains its all adherent point. So, A is closed.
Note: In remark 2, we CANNOT use the relation<br />
fx gx<br />
since the difference "" are not necessarily defined on the metric space T, d 2 .<br />
4.23 Given a function f : R R, define two sets A and B in R 2 as follows:<br />
A x, y : y fx,<br />
B x, y : y fx.<br />
and Prove that f is continuous on R if, and only if, both A and B are open subsets of R 2 .<br />
Proof: () Suppose that f is continuous on R. Leta, b A, then fa b. Sincef is<br />
continuous at a, thengiven fab 0, there exists a 0 such that as<br />
2<br />
|x a| , wehave<br />
fa b<br />
fa fx fa . *<br />
2<br />
Consider x, y Ba, b; , then |x a| 2 |y b| 2 2 which implies that<br />
1. |x a| fx fa b by (*) and<br />
2<br />
2. |y b| y b b fa b .<br />
2<br />
Hence, we have fx y. Thatis,Ba, b; A. So,A is open since every point of A is<br />
interior. Similarly for B.<br />
() Suppose that A and B are open in R 2 . Trivially, a, fa /2 : p 1 A, and<br />
a, fa /2 : p 2 B. SinceA and B are open in R 2 , there exists a /2 0 such<br />
that<br />
Bp 1 , A and Bp 2 , B.<br />
Hence, if x, y Bp 1 , , then<br />
x a 2 y fa /2 2 2 and y fx.<br />
So, it implies that<br />
|x a| , |y fa /2| , andy fx.<br />
Hence, as |x a| , wehave<br />
y fa /2<br />
fa /2 y fx<br />
fa y fx<br />
fa fx. **<br />
<strong>And</strong> if x, y Bp 2 , , then<br />
x a 2 y fa /2 2 2 and y fx.<br />
So, it implies that<br />
|x a| , |y fa /2| , andy fx.<br />
Hence, as |x a| , wehave<br />
fx y fa /2 fa . ***<br />
So, given 0, there exists a 0 such that as |x a| , we have by (**) and (***)<br />
fa fx fa .<br />
That is, f is continuous at a. Sincea is arbitrary, we know that f is continuous on R.<br />
4.24 Let f be defined and bounded on a compact interval S in R. IfT S, the
number<br />
f T supfx fy : x, y T<br />
is called the oscillation (or span) of f on T. Ifx S, the oscillation of f at x is defined to<br />
be the number<br />
f x lim <br />
h0<br />
<br />
fBx; h S.<br />
Prove that this limit always exists and that f x 0if,andonlyif,f is continuous at x.<br />
Proof: 1. Note that since f is bounded, say |fx| M for all x, wehave<br />
|fx fy| 2M for all x, y S. So, f T, the oscillation of f on any subset T of S,<br />
exists. In addition, we define gh f Bx; h S. Note that if T 1 T 2 S, wehave<br />
f T 1 f T 2 . Hence, the oscillation of f at x, f x lim h0<br />
gh g0 since g is<br />
an increasing function. That is, the limit of f Bx; h S always exists as h 0 .<br />
2. Suppose that f x 0, then given 0, there exists a 0 such that as<br />
h 0, , wehave<br />
|gh| gh f Bx; h S /2.<br />
That is, as h 0, , wehave<br />
supft fs : t, s Bx; h S /2<br />
which implies that<br />
/2 ft fx /2 as t x , x S.<br />
So, as t x , x S. wehave<br />
|ft fx| .<br />
That is, f is continuous at x.<br />
Suppose that f is continous at x, thengiven 0, there exists a 0 such that as<br />
t x , x S, wehave<br />
|ft fx| /3.<br />
So, as t, s x , x S, wehave<br />
|ft fs| |ft fx| |fx fs| /3 /3 2/3<br />
which implies that<br />
supt fs : t, s x , x S 2/3 .<br />
So, as h 0, , wehave<br />
f Bx; h S supt fs : t, s x , x S .<br />
Hence, the oscillation of f at x, f x 0.<br />
Remark: 1. <strong>The</strong> compactness of S is not used here, we will see the advantage of the<br />
oscillation of f in text book, <strong>The</strong>orem 7.48, in page 171. (On Lebesgue’s Criterion for<br />
Riemann-Integrability.)<br />
2. One of advantage of the oscillation of f is to show the statement: Let f be defined on<br />
a, b Prove that a bounded f does NOT have the properties:<br />
f is continuous on Q a, b, and discontinuous on R Q a, b.<br />
Proof: Since f x 0if,andonlyif,f is continuous at x, we know that f r 0for<br />
r R Q a, b. DefineJ 1/n r : f r 1/n, then by hypothesis, we know that<br />
<br />
n1 J 1/n R Q a, b. It is easy to show that J 1/n is closed in a, b. Hence,<br />
intclJ 1/n intJ 1/n for all n N. It means that J 1/n is nowhere dense for all<br />
n N. Hence,
a, b <br />
n1 J 1/n Q a, b<br />
is of the firse category which is absurb since every complete metric space is of the second<br />
category. So, this f cannot exist.<br />
Note: 1 <strong>The</strong> Boundedness of f can be removed since we we can accept the concept<br />
0.<br />
2. (J 1/n is closed in a, b) Given an accumulation point x of J 1/n ,ifx J 1/n ,wehave<br />
f x 1/n. So, there exists a 1 ball Bx such that f Bx a, b 1/n. Thus, no<br />
points of Bx can belong to J 1/n , contradicting that x is an accumulation point of J 1/n .<br />
Hence, x J 1/n and J 1/n is closed.<br />
3. (Definition of a nowhere dense set) In a metric space M, d, letA be a subset of M,<br />
we say A is nowhere dense in M if, and only if A contains no balls of M, ( intA ).<br />
4. (Definition of a set of the first category and of the second category) AsetA in a<br />
metric space M is of the first category if, and only if, A is the union of a countable number<br />
of nowhere dense sets. A set B is of the second category if, and only if, B is not of the first<br />
category.<br />
5. (<strong>The</strong>orem) A complete metric space is of the second category.<br />
We write another important theorem about a set of the second category below.<br />
(Baire Category <strong>The</strong>orem) A nonempty open set in a complete metric space is of the<br />
second category.<br />
6. In the notes 3,4 and 5, the reader can see the reference, A First Course in <strong>Real</strong><br />
Analysis written by M. H. Protter and C. B. Morrey, in pages 375-377.<br />
4.25 Let f be continuous on a compact interval a, b. Suppose that f has a local<br />
maximum at x 1 and a local maximum at x 2 . Show that there must be a third point between<br />
x 1 and x 2 where f has a local minimum.<br />
Note.Tosaythatf has a local maximum at x 1 means that there is an 1 ball Bx 1 such<br />
that fx fx 1 for all x in Bx 1 a, b. Local minimum is similarly defined.<br />
Proof: Let x 2 x 1 . Suppose NOT, i.e., no points on x 1 , x 2 can be a local minimum<br />
of f. Sincef is continuous on x 1 , x 2 , then inffx : x x 1 , x 2 fx 1 or fx 2 by<br />
hypothesis. We consider two cases as follows:<br />
(1) If inffx : x x 1 , x 2 fx 1 , then<br />
(i) fx has a local maximum at x 1 and<br />
(ii) fx fx 1 for all x x 1 , x 2 .<br />
By (i), there exists a 0 such that x x 1 , x 1 x 1 , x 2 ,wehave<br />
(iii) fx fx 1 .<br />
So, by (ii) and (iii), as x x 1 , x 1 , wehave<br />
fx fx 1 <br />
which contradicts the hypothesis that no points on x 1 , x 2 can be a local minimum of f.<br />
(2) If inffx : x x 1 , x 2 fx 1 , it is similar, we omit it.<br />
Hence, from (1) and (2), we have there has a third point between x 1 and x 2 where f has<br />
a local minimum.<br />
4.26 Let f be a real-valued function, continuous on 0, 1, with the following property:<br />
For every real y, either there is no x in 0, 1 for which fx y or there is exactly one such<br />
x. Prove that f is strictly monotonic on 0, 1.
Proof: Since the hypothesis says that f is one-to-one, then by <strong>The</strong>orem*, we know that f<br />
is trictly monotonic on 0, 1.<br />
Remark: (<strong>The</strong>orem*) Under assumption of continuity on a compact interval, 1-1 is<br />
equivalent to being strictly monotonic. We will prove it in Exercise 4.62.<br />
4.27 Let f be a function defined on 0, 1 with the following property: For every real<br />
number y, either there is no x in 0, 1 for which fx y or there are exactly two values of<br />
x in 0, 1 for which fx y.<br />
(a) Prove that f cannot be continuous on 0, 1.<br />
Proof: Assume that f is continuous on 0, 1, and thus consider max x0,1 fx and<br />
min x0,1 fx. <strong>The</strong>n by hypothesis, there exist exactly two values a 1 a 2 0, 1 such<br />
that fa 1 fa 2 max x0,1 fx, and there exist exactly two values b 1 b 2 0, 1<br />
such that fb 1 fb 2 min x0,1 fx.<br />
Claim that a 1 0anda 2 1. Suppose NOT, then there exists at least one belonging<br />
to 0, 1. Without loss of generality, say a 1 0, 1. Sincef has maximum at a 1 0, 1<br />
and a 2 0, 1, we can find three points p 1 , p 2 ,andp 3 such that<br />
1. p 1 a 1 p 2 p 3 a 2 ,<br />
2. fp 1 fa 1 , fp 2 fa 1 ,andfp 3 fa 2 .<br />
Since fa 1 fa 2 , we choose a real number r so that<br />
fp 1 r fa 1 r fq 1 ,whereq 1 p 1 , a 1 by continuity of f.<br />
fp 2 r fa 1 r fq 2 ,whereq 2 a 1 , p 2 by continuity of f.<br />
fp 3 r fa 2 r fq 3 ,whereq 3 p 3 , a 2 by continuity of f.<br />
which contradicts the hypothesis that for every real number y, there are exactly two values<br />
of x in 0, 1 for which fx y. Hence, we know that a 1 0anda 2 1. Similarly, we<br />
also have b 1 0andb 2 1.<br />
So, max x0,1 fx min x0,1 fx which implies that f is constant. It is impossible.<br />
Hence, such f does not exist. That is, f is not continuous on 0, 1.<br />
(b) Construct a function f which has the above property.<br />
Proof: Consider 0, 1 Q c 0, 1 Q 0, 1, and write<br />
Q 0, 1 x 1 , x 2 ,...,x n ,.... Define<br />
1. fx 2n1 fx 2n n,<br />
2. fx x if x 0, 1/2 Q c ,<br />
3. fx 1 x if x 1/2, 1 Q c .<br />
<strong>The</strong>n if x y, then it is clear that fx fy. Thatis,f is well-defined. <strong>And</strong> from<br />
construction, we know that the function defined on 0, 1 with the following property: For<br />
every real number y, either there is no x in 0, 1 for which fx y or there are exactly<br />
two values of x in 0, 1 for which fx y.<br />
Remark: x : f is discontinuous at x 0, 1. Givena 0, 1. Note that since<br />
fx N for all x Q 0, 1 and Q is dense in R, forany1ball Ba; r Q 0, 1,<br />
there is always a rational number y Ba; r Q 0, 1 such that |fy fa| 1.<br />
(c) Prove that any function with this property has infinite many discontinuities on<br />
0, 1.<br />
Proof: In order to make the proof clear, property A of f means that
for every real number y, either there is no x in 0, 1 for which fx y or<br />
there are exactly two values of x in 0, 1 for which fx y<br />
Assume that there exist a finite many numbers of discontinuities of f, say these points<br />
x 1 ,...,x n . By property A, there exists a unique y i such that fx i fy i for 1 i n.<br />
Note that the number of the set<br />
S : x 1 ,...,x n y 1 ,...,y n x : fx f0, andfx f1 is even, say 2m<br />
We remove these points from S, and thus we have 2m 1 subintervals, say I j ,<br />
1 j 2m 1. Consider the local extremum in every I j ,1 j 2m 1 and note that<br />
every subinterval I j ,1 j 2m 1, has at most finite many numbers of local extremum,<br />
say # t I j : fx is the local extremum t j j<br />
1 ,..,t pj p j . <strong>And</strong> by property A,<br />
there exists a unique s j j<br />
j<br />
k such that f t k f s k for 1 k p j . We again remove these<br />
points, and thus we have removed even number of points. <strong>And</strong> odd number of open<br />
intervals is left, call the odd number 2q 1. Note that since the function f is monotonic in<br />
every open interval left, R l ,1 l 2q 1, the image of f on these open interval is also an<br />
open interval. If R a R b , sayR a a 1 , a 1 and R b b 1 , b 2 with (without loss of<br />
generality) a 1 b 1 a 2 b 2 , then<br />
R a R b by property A.<br />
(Otherwise, b 1 is only point such that fx fb 1 , which contradicts property A.) Note<br />
that given any R a , there must has one and only one R b such that R a R b . However, we<br />
have 2q 1 open intervals is left, it is impossible. Hence, we know that f has infinite many<br />
discontinuities on 0, 1.<br />
4.28 In each case, give an example of a real-valued function f, continuous on S and<br />
such that fS T, or else explain why there can be no such f :<br />
(a) S 0, 1, T 0, 1.<br />
Solution: Let<br />
fx <br />
2x if x 0, 1/2,<br />
1ifx 1/2, 1.<br />
(b) S 0, 1, T 0, 1 1, 2.<br />
Solution: NO! Since a continuous functions sends a connected set to a connected set.<br />
However, in this case, S is connected and T is not connected.<br />
(c) S R 1 , T the set of rational numbers.<br />
Solution: NO! Since a continuous functions sends a connected set to a connected set.<br />
However, in this case, S is connected and T is not connected.<br />
(d) S 0, 1 2, 3, T 0, 1.<br />
Solution: Let<br />
(e) S 0, 1 0, 1, T R 2 .<br />
fx <br />
0ifx 0, 1,<br />
1ifx 2, 3.<br />
Solution: NO! Since a continuous functions sends a compact set to a compact set.<br />
However, in this case, S is compact and T is not compact.
(f) S 0, 1 0, 1, T 0, 1 0, 1.<br />
Solution: NO! Since a continuous functions sends a compact set to a compact set.<br />
However, in this case, S is compact and T is not compact.<br />
(g) S 0, 1 0, 1, T R 2 .<br />
Solution: Let<br />
fx, y cot x,coty.<br />
Remark: 1. <strong>The</strong>re is some important theorems. We write them as follows.<br />
(<strong>The</strong>orem A) Letf : S, d s T, d T be continuous. If X is a compact subset of S,<br />
then fX is a compact subset of T.<br />
(<strong>The</strong>orem B) Letf : S, d s T, d T be continuous. If X is a connected subset of S,<br />
then fX is a connected subset of T.<br />
2. In (g), the key to the example is to find a continuous function f : 0, 1 R which is<br />
onto.<br />
Supplement on Continuity of real valued functions<br />
Exercise Suppose that fx : 0, R, is continuous with a fx b for all<br />
x 0, , and for any real y, either there is no x in 0, for which fx y or<br />
there are finitely many x in 0, for which fx y. Prove that lim x fx exists.<br />
Proof: For convenience, we say property A, it means that for any real y, either<br />
there is no x in 0, for which fx y or there are finitely many x in 0, for<br />
which fx y.<br />
We partition a, b into n subintervals. <strong>The</strong>n, by continuity and property A, asx<br />
is large enough, fx is lying in one and only one subinterval. Given 0, there<br />
exists N such that 2/N . For this N, we partition a, b into N subintervals, then<br />
there is a M 0 such that as x, y M<br />
|fx fy| 2/N .<br />
So, lim x fx exists.<br />
Exercise Suppose that fx : 0, 1 R is continunous with f0 f1 0. Prove that<br />
(a) there exist two points x 1 and x 2 such that as |x 1 x 2 | 1/n, wehave<br />
fx 1 fx 2 0 for all n. In this case, we call 1/n the length of horizontal strings.<br />
Proof: Define a new function gx fx 1 n fx : 0, 1 1 n . Claim that<br />
there exists p 0, 1 1 n such that gp 0. Suppose NOT, byIntermediate<br />
Value <strong>The</strong>orem, without loss of generality, let gx 0, then<br />
g0 g 1 n ...g 1 1 n f1 0<br />
which is absurb. Hence, we know that there exists p 0, 1 1 n such that<br />
gp 0. That is,<br />
f p 1 n fp.<br />
So,wehave1/n as the length of horizontal strings.<br />
(b) Could you show that there exists 2/3 as the length of horizontal strings<br />
Proof: <strong>The</strong> horizontal strings does not exist, for example,
fx <br />
x, ifx 0, 1 4<br />
x 1 ,ifx 1 , 3 .<br />
2 4 4<br />
x 1, if x 3 ,1 4<br />
Exercise Suppose that fx : a, b R is a continuous and non-constant function. Prove<br />
that the function f cannot have any small periods.<br />
Proof: Say f is continuous at q a, b, and by hypothesis that f is<br />
non-constant, there is a point p a, b such that |fq fp| : M 0. Since f is<br />
continuous at q, thengiven M, there is a 0 such that as<br />
x q , q a, b, wehave<br />
|fx fq| M. *<br />
If f has any small periods, then in the set q , q a, b, there is a point<br />
r q , q a, b such that fr fp. It contradicts to (*). Hence, the<br />
function f cannot have any small periods.<br />
Remark 1. <strong>The</strong>re is a function with any small periods.<br />
Solution:<strong>The</strong> example is Dirichlet function,<br />
0, if x Q<br />
fx <br />
c<br />
1, if x Q .<br />
Since fx q fx, for any rational q, we know that f has any small periods.<br />
2. Prove that there cannot have a non-constant continuous function which has<br />
two period p, andq such that q/p is irrational.<br />
Proof: Since q/p is irrational, there is a sequence qn<br />
p n<br />
Q such that<br />
q n<br />
p n<br />
q p 1 p<br />
<br />
p2 |pq n qp n| <br />
n<br />
p n<br />
0asn .<br />
So, f has any small periods, by this exercise, we know that this f cannot a<br />
non-constant continuous function.<br />
Note: <strong>The</strong> inequality is important; the reader should kepp it in mind. <strong>The</strong>re are<br />
many ways to prove this inequality, we metion two methods without proofs. <strong>The</strong><br />
reader can find the proofs in the following references.<br />
(1) An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s written by G.H. Hardy and<br />
E.M. Wright, charpter X, pp 137-138.<br />
(2) In the text book, exercise 1.15 and 1.16, pp 26.<br />
3. Suppose that fx is differentiable on R prove that if f has any small periods,<br />
then f is constant.<br />
Proof: Givenc R, and consider<br />
fc p n fc<br />
p n<br />
0 for all n.<br />
where p n is a sequence of periods of function such that p n 0. Hence, by<br />
differentiability of f, we know that f c 0. Since c is arbitrary, we know that<br />
f x 0onR. Hence, f is constant.<br />
Continuity in metric spaces
In Exercises 4.29 through 4.33, we assume that f : S T is a function from one metric<br />
space S, d S to another T, d T .<br />
4.29 Prove that f is continuous on S if, and only if,<br />
f 1 intB intf 1 B for every subset B of T.<br />
Proof: () Suppose that f is continuous on S, and let B be a subset of T. Since<br />
intB B, wehavef 1 intB f 1 B. Note that f 1 intB is open since a pull back of<br />
an open set under a continuous function is open. Hence, we have<br />
intf 1 intB f 1 intB intf 1 B.<br />
That is, f 1 intB intf 1 B for every subset B of T.<br />
() Suppose that f 1 intB intf 1 B for every subset B of T. Given an open subset<br />
U T, i.e., intU U, sowehave<br />
f 1 U f 1 intU intf 1 U.<br />
In addition, intf 1 U f 1 U by the fact, for any set A, intA is a subset of A. So,f is<br />
continuous on S.<br />
4.30 Prove that f is continuous on S if, and only if,<br />
fclA clfA for every subset A of S.<br />
Proof: () Suppose that f is continuous on S, and let A be a subset of S. Since<br />
fA clfA, then A f 1 fA f 1 clfA. Note that f 1 clfA is closed<br />
since a pull back of a closed set under a continuous function is closed. Hence, we have<br />
clA clf 1 clfA f 1 clfA<br />
which implies that<br />
fclA ff 1 clfA clfA.<br />
() Suppose that fclA clfA for every subset A of S. Given a closed subset<br />
C T, and consider f 1 C as follows. Define f 1 C A, then<br />
fclf 1 C fclA<br />
clfA clff 1 C<br />
clC C since C is closed.<br />
So,wehaveby(fclA C)<br />
clA f 1 fclA f 1 C A<br />
which implies that A f 1 C is closed set. So, f is continuous on S.<br />
4.31 Prove that f is continuous on S if, and only if, f is continuous on every compact<br />
subset of S. Hint. If x n p in S, the set p, x 1 , x 2 ,... is compact.<br />
Proof: () Suppose that f is continuous on S, then it is clear that f is continuous on<br />
every compact subset of S.<br />
() Suppose that f is continuous on every compact subset of S, Givenp S, we<br />
consider two cases.<br />
(1) p is an isolated point of S, then f is automatically continuous at p.<br />
(2) p is not an isolated point of S, thatis,p is an accumulation point p of S, then there<br />
exists a sequence x n S with x n p. Note that the set p, x 1 , x 2 ,... is compact, so we<br />
know that f is continuous at p. Sincep is arbitrary, we know that f is continuous on S.<br />
Remark: If x n p in S, the set p, x 1 , x 2 ,... is compact. <strong>The</strong> fact is immediately
from the statement that every infinite subset p, x 1 , x 2 ,... of has an accumulation point in<br />
p, x 1 , x 2 ,....<br />
4.32 A function f : S T is called a closed mapping on S if the image fA is closed<br />
in T for every closed subset A of S. Prove that f is continuous and closed on S if, and only<br />
if,<br />
fclA clfA for every subset A of S.<br />
Proof: () Suppose that f is continuous and closed on S, and let A be a subset of S.<br />
Since A clA, wehavefA fclA. So, we have<br />
clfA clfclA fclA since f is closed. *<br />
In addition, since fA clfA, wehaveA f 1 fA f 1 clfA. Note that<br />
f 1 clfA is closed since f is continuous. So, we have<br />
clA clf 1 clfA f 1 clfA<br />
which implies that<br />
fclA ff 1 clfA clfA. **<br />
From (*) and (**), we know that fclA clfA for every subset A of S.<br />
() Suppose that fclA clfA for every subset A of S. Gvien a closed subset C<br />
of S, i.e., clC C, then we have<br />
fC fclC clfC.<br />
So, we have fC is closed. That is, f is closed. Given any closed subset B of T, i.e.,<br />
clB B, we want to show that f 1 B is closed. Since f 1 B : A S, wehave<br />
fclf 1 B fclA clfA clff 1 B clB B<br />
which implies that<br />
fclf 1 B B clf 1 B f 1 fclf 1 B f 1 B.<br />
That is, we have clf 1 B f 1 B. So,f 1 B is closed. Hence, f is continuous on S.<br />
4.33 Give an example of a continuous f and a Cauchy sequence x n in some metric<br />
space S for which fx n is not a Cauchy sequence in T.<br />
Solution: Let S 0, 1, x n 1/n for all n N, andf 1/x : S R. <strong>The</strong>n it is clear<br />
that f is continous on S, andx n is a Cauchy sequence on S. In addition, Trivially,<br />
fx n n is not a Cauchy sequence.<br />
Remark: <strong>The</strong> reader may compare the exercise with the Exercise 4.54.<br />
4.34 Prove that the interval 1, 1 in R 1 is homeomorphic to R 1 . This shows that<br />
neither boundedness nor completeness is a topological property.<br />
Proof: Since fx tan x : 1, 1 R is bijection and continuous, and its<br />
2<br />
converse function f 1 x arctanx : R 1, 1. Hence, we know that f is a Topologic<br />
mapping. (Or say f is a homeomorphism). Hence, 1, 1 is homeomorphic to R 1 .<br />
Remark: A function f is called a bijection if, and only if, f is 1-1 and onto.<br />
4.35 Section 9.7 contains an example of a function f, continuous on 0, 1, with<br />
f0, 1 0, 1 0, 1. Prove that no such f can be one-to-one on 0, 1.<br />
Proof: By section 9.7, let f : 0, 1 0, 1 0, 1 be an onto and continuous function.<br />
If f is 1-1, then so is its converse function f 1 . Note that since f is a 1-1 and continous<br />
function defined on a compact set 0, 1, then its converse function f 1 is also a continous
function. Since f0, 1 0, 1 0, 1, we have the domain of f 1 is 0, 1 0, 1 which<br />
is connected. Choose a special point y 0, 1 0, 1 so that f 1 y : x 0, 1.<br />
Consider a continous function g f 1 | 0,10,1y , then<br />
g : 0, 1 0, 1 y 0, x x,1 which is continous. However, it is impossible<br />
since 0, 1 0, 1 y is connected but 0, x x,1 is not connected. So, such f cannot<br />
exist.<br />
Connectedness<br />
4.36 Prove that a metric space S is disconnected if, and only if there is a nonempty<br />
subset A of S, A S, which is both open and closed in S.<br />
Proof: () Suppose that S is disconnected, then there exist two subset A, B in S such<br />
that<br />
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.<br />
Note that since A, B are open in S ,wehaveA S B, B S A are closed in S. So,ifS<br />
is disconnected, then there is a nonempty subset A of S, A S, which is both open and<br />
closed in S.<br />
() Suppose that there is a nonempty subset A of S, A S, which is both open and<br />
closed in S. <strong>The</strong>n we have S A : B is nonempty and B is open in S. Hence, we have two<br />
sets A, B in S such that<br />
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.<br />
That is, S is disconnected.<br />
4.37 Prove that a metric space S is connected if, and only if the only subsets of S which<br />
are both open and closed in S are empty set and S itself.<br />
Proof: () Suppose that S is connected. If there exists a subset A of S such that<br />
1. A , 2.A is a proper subset of S, 3.A is open and closed in S,<br />
then let B S A, wehave<br />
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.<br />
It is impossible since S is connected. So, this A cannot exist. That is, the only subsets of S<br />
which are both open and closed in S are empty set and S itself.<br />
() Suppose that the only subsets of S which are both open and closed in S are empty<br />
set and S itself. If S is disconnected, then we have two sets A, B in S such that<br />
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.<br />
It contradicts the hypothesis that the only subsets of S which are both open and closed in S<br />
are empty set and S itself.<br />
Hence, we have proved that S is connected if, and only if the only subsets of S which<br />
are both open and closed in S are empty set and S itself.<br />
4.38 Prove that the only connected subsets of R are<br />
(a) the empty set,<br />
(b) sets consisting of a single point, and<br />
(c) intervals (open, closed, half-open, or infinite).<br />
Proof: Let S be a connected subset of R. Denote the symbol #A to be the number of<br />
elements in a set A. We consider three cases as follows. (a) #S 0, (b) #S 1, (c)<br />
#S 1.<br />
For case (a), it means that S , and for case (b), it means that S consists of a single<br />
point. It remains to consider the case (c). Note that since #S 1, we have inf S sup S.
Since S R, wehaveS inf S,supS. (Note that we accept that inf S or<br />
sup S .) If S is not an interval, then there exists x inf S,supS such that x S.<br />
(Otherwise, inf S,supS S which implies that S is an interval.) <strong>The</strong>n we have<br />
1. , x S : A is open in S<br />
2. x, S : B is open in S<br />
3. A B S.<br />
Claim that both A and B are not empty. Asume that A is empty, then every s S, wehave<br />
s x inf S. By the definition of infimum, it is impossible. So, A is not empty. Similarly<br />
for B. Hence, we have proved that S is disconnected, a contradiction. That is, S is an<br />
interval.<br />
Remark: 1. We note that any interval in R is connected. It is immediate from Exercise<br />
4.44. But we give another proof as follows. Suppose there exists an interval S is not<br />
connceted, then there exist two subsets A and B such that<br />
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S.<br />
Since A and B , we choose a A and b B, and let a b. Consider<br />
c : supA a, b.<br />
Note that c clA A implies that c B. Hence, we have a c b. In addition,<br />
c B clB, then there exists a B S c; B . Choose<br />
d B S c; c , c S so that<br />
1. c d b and 2. d B.<br />
<strong>The</strong>n d A. (Otherwise, it contradicts c supA a, b. Note that<br />
d a, b S A B which implies that d A or d B. We reach a contradiction since<br />
d A and d B. Hence, we have proved that any interval in R is connected.<br />
2. Here is an application. Is there a continuous function f : R R such that<br />
fQ Q c ,andfQ c Q <br />
Ans: NO! If such f exists, then both fQ and fQ c are countable. Hence, fR is<br />
countable. In addition, fR is connected. Since fR contains rationals and irrationals, we<br />
know fR is an interval which implies that fR is uncountable, a cotradiction. Hence, such<br />
f does not exist.<br />
4.39 Let X be a connected subset of a metric space S. LetY be a subset of S such that<br />
X Y clX, whereclX is the closure of X. Prove that Y is also connected. In<br />
particular, this shows that clX is connected.<br />
Proof: Given a two valued function f on Y, we know that f is also a two valued<br />
function on X. Hence, f is constant on X, (without loss of generality) say f 0onX.<br />
Consider p Y X, it ,means that p is an accumulation point of X. <strong>The</strong>n there exists a<br />
dequence x n X such that x n p. Note that fx n 0 for all n. So, we have by<br />
continuity of f on Y,<br />
fp f lim n<br />
x n<br />
lim n<br />
fx n 0.<br />
Hence, we have f is constant 0 on Y. Thatis,Y is conneceted. In particular, clX is<br />
connected.<br />
Remark: Of course, we can use definition of a connected set to show the exercise. But,<br />
it is too tedious to write. However, it is a good practice to use definition to show it. <strong>The</strong><br />
reader may give it a try as a challenge.
4.40 If x is a point in a metric space S, letUx be the component of S containing x.<br />
Prove that Ux is closed in S.<br />
Proof: Let p be an accumulation point of Ux. Letf be a two valued function defined<br />
on Ux p, then f is a two valud function defined on Ux. SinceUx is a component<br />
of S containing x, then Ux is connected. That is, f is constant on Ux, (without loss of<br />
generality) say f 0onUx. <strong>And</strong> since p is an accumulation point of Ux, there exists a<br />
sequence x n Ux such that x n p. Note that fx n 0 for all n. So, we have by<br />
continuity of f on Ux p,<br />
fp f lim n<br />
x n<br />
lim n<br />
fx n 0.<br />
So, Ux p is a connected set containing x. SinceUx is a component of S containing<br />
x, wehaveUx p Ux which implies that p Ux. Hence, Ux contains its all<br />
accumulation point. That is, Ux is closed in S.<br />
4.41 Let S be an open subset of R. By <strong>The</strong>orem 3.11, S is the union of a countable<br />
disjoint collection of open intervals in R. Prove that each of these open intervals is a<br />
component of the metric subspace S. Explain why this does not contradict Exercise 4.40.<br />
Proof: By <strong>The</strong>orem 3.11, S <br />
n1 I n ,whereI i is open in R and I i I j if i j.<br />
Assume that there exists a I m such that I m is not a component T of S. <strong>The</strong>n T I m is not<br />
empty. So, there exists x T I m and x I n for some n. Note that the component Ux is<br />
the union of all connected subsets containing x, then we have<br />
T I n Ux. *<br />
In addition,<br />
Ux T **<br />
since T is a component containing x. Hence, by (*) and (**), we have I n T. So,<br />
I m I n T. SinceT is connected in R 1 , T itself is an interval. So, intT is still an interval<br />
which is open and containing I m I n . It contradicts the definition of component interval.<br />
Hence, each of these open intervals is a component of the metric subspace S.<br />
Since these open intervals is open relative to R, not S, this does not contradict Exercise<br />
4.40.<br />
4.42 Given a compact S in R m with the following property: For every pair of points a<br />
and b in S and for every 0 there exists a finite set of points x 0 , x 1 ,...,x n in S with<br />
x 0 a and x n b such that<br />
x k x k1 for k 1,2,..,n.<br />
Prove or disprove: S is connected.<br />
Proof: Suppose that S is disconnected, then there exist two subsets A and B such that<br />
1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S. *<br />
Since A and B , we choose a A, andb A and thus given 1, then by<br />
hypothesis, we can find two points a 1 A, andb 1 B such that a 1 b 1 1. For a 1 ,<br />
and b 1 ,given 1/2, then by hypothesis, we can find two points a 2 A, andb 2 B<br />
such that a 2 b 2 1/2. Continuous the steps, we finally have two sequence a n A<br />
and b n B such that a n b n 1/n for all n. Sincea n A, andb n B, we<br />
have a n S and b n S by S A B. Hence, there exist two subsequence<br />
a nk A and b nk B such that a nk x, andb nk y, wherex, y S since S is<br />
compact. In addition, since A is closed in S, andB is closed in S, wehavex A and<br />
y B. On the other hand, since a n b n 1/n for all n, wehavex y. Thatis,
A B which contradicts (*)-3. Hence, we have prove that S is connected.<br />
Remark: We given another proof by the method of two valued function as follows. Let<br />
f be a two valued function defined on S, and choose any two points a, b S. If we can<br />
show that fa fb, we have proved that f is a constant which implies that S is<br />
connected. Since f is a continuous function defined on a compact set S, then f is uniformly<br />
on S. Thus, given 1 0, there exists a 0 such that as x y , x, y S, we<br />
have |fx fy| 1 fx fy. Hence, for this , there exists a finite set of<br />
points x 0 , x 1 ,...,x n in S with x 0 a and x n b such that<br />
x k x k1 for k 1,2,..,n.<br />
So, we have fa fx 0 fx 1 ... fx n fb.<br />
4.43 Prove that a metric space S is connected if, and only if, every nonempty proper<br />
subset of S has a nonempty boundary.<br />
Proof: () Suppose that S is connected, and if there exists a nonempty proper subset U<br />
of S such that U , thenletB clS U, wehave(defineclU A<br />
1. A . B since S U ,<br />
2. A B clU clS U U S U S<br />
S A B,<br />
3. A B clU clS U U ,<br />
and<br />
4. Both A and B are closed in S Both A and B are open in S.<br />
Hence, S is disconnected. That is, if S is connected, then every nonempty proper subset of<br />
S has a nonempty boundary.<br />
() Suppose that every nonempty proper subset of S has a nonempty boundary. If S is<br />
disconnected, then there exist two subsets A and B such that<br />
1. A, B are closed in S, 2.A and B , 3.A B , and 4. A B S.<br />
<strong>The</strong>n for this A, A is a nonempty proper subset of S with (clA A, andclB B)<br />
A clA clS A clA clB A B <br />
which contradicts the hypothesis that every nonempty proper subset of S has a nonempty<br />
boundary. So, S is connected.<br />
4.44. Prove that every convex subset of R n is connected.<br />
Proof: Given a convex subset S of R n , and since for any pair of points a, b, the set<br />
1 a b :0 1 : T S, i.e., g : 0, 1 T by g 1 a b is a<br />
continuous function such that g0 a, andg1 b. So,S is path-connected. It implies<br />
that S is connected.<br />
Remark: 1. In the exercise, it tells us that every n ball is connected. (In fact, every<br />
n ball is path-connected.) In particular, as n 1, any interval (open, closed, half-open, or<br />
infinite) in R is connected. For n 2, any disk (open, closed, or not) in R 2 is connected.<br />
2. Here is a good exercise on the fact that a path-connected set is connected. Given<br />
0, 1 0, 1 : S, andifT is a countable subset of S. Prove that S T is connected. (In<br />
fact, S T is path-connected.)<br />
Proof: Given any two points a and b in S T, then consider the vertical line L passing<br />
through the middle point a b/2. Let A x : x L S, and consider the lines form<br />
a to A, and from b to A. Note that A is uncountable, and two such lines (form a to A, and
from b to A) are disjoint. So, if every line contains a point of T, then it leads us to get T is<br />
uncountable. However, T is countable. So, it has some line (form a to A, and from b to A)<br />
is in S T. So, it means that S T is path-connected. So, S T is connected.<br />
4.45 Given a function f : R n R m which is 1-1 and continuous on R n .IfA is open<br />
and disconnected in R n , prove that fA is open and disconnected in fR n .<br />
Proof: <strong>The</strong>exerciseiswrong. <strong>The</strong>re is a counter-example. Let f : R R 2<br />
f <br />
cos 2x<br />
1x<br />
cos 2x<br />
1x<br />
2<br />
2<br />
2x<br />
,1 sin if x 0<br />
1x 2<br />
2x<br />
, 1 sin if x 0<br />
1x 2<br />
Remark: If we restrict n, m 1, the conclusion holds. That is, Let f : R R be<br />
continuous and 1-1. If A is open and disconnected, then so is fA.<br />
Proof: In order to show this, it suffices to show that f maps an open interval I to<br />
another open interval. Since f is continuous on I, andI is connected, fI is connected. It<br />
implies that fI is an interval. Trivially, there is no point x in I such that fx equals the<br />
endpoints of fI. Hence, we know that fI is an open interval.<br />
Supplement: Here are two exercises on Homeomorphism to make the reader get more<br />
and feel something.<br />
1. Let f : E R R. Ifx, fx : x E is compact, then f is uniformly continuous<br />
on E.<br />
Proof: Let x, fx : x E S, and thus define gx Ix x, fx : E S.<br />
Claim that g is continuous on E. Consider h : S E by hx, fx x. Trivially, h is 1-1,<br />
continuous on a compact set S. So, its inverse function g is 1-1 and continuous on a<br />
compact set E. <strong>The</strong> claim has proved.<br />
Since g is continuous on E, we know that f is continuous on a compact set E. Hence, f<br />
is uniformly continuous on E.<br />
Note: <strong>The</strong> question in Supplement 1, there has another proof by the method of<br />
contradiction, and use the property of compactness. We omit it.<br />
2. Let f : 0, 1 R. Ifx, fx : x 0, 1 is path-connected, then f is continuous<br />
on 0, 1.<br />
Proof: Let a 0, 1, then there is a compact interval a a 1 , a 2 0, 1. Claim<br />
that the set<br />
x, fx : x a 1 , a 2 : S is compact.<br />
Since S is path-connected, there is a continuous function g : 0, 1 S such that<br />
g0 a 1 , fa 1 and g1 a 2 , fa 2 . If we can show g0, 1 S, wehaveshown<br />
that S is compact. Consider h : S R by hx, fx x; h is clearly continuous on S. So,<br />
the composite function h g : 0, 1 R is also continuous. Note that h g0 a 1 ,and<br />
h g1 a 2 , and the range of h g is connected. So, a 1 , a 2 hg0, 1. Hence,<br />
g0, 1 S. We have proved the claim and by Supplement 1, we know that f is<br />
continuous at a. Sincea is arbitrary, we know that f is continuous on 0, 1.<br />
Note: <strong>The</strong> question in Supplement 2, there has another proof directly by definition of<br />
continuity. We omit the proof.<br />
4.46 Let A x, y :0 x 1, y sin 1/x, B x, y : y 0, 1 x 0,<br />
and let S A B. Prove that S is connected but not arcwise conneceted. (See Fig. 4.5,
Section 4.18.)<br />
Proof: Let f be a two valued function defined on S. SinceA, andB are connected in S,<br />
then we have<br />
fA a, andfB b, wherea, b 0, 1.<br />
Given a sequence x n A with x n 0, 0, then we have<br />
a lim n<br />
fx n f limx n by continuity of f at 0<br />
n0<br />
f0, 0<br />
b.<br />
So, we have f is a constant. That is, S is connected.<br />
Assume that S is arcwise connected, then there exists a continuous function<br />
g : 0, 1 S such that g0 0, 0 and g1 1, sin 1. Given 1/2, there exists a<br />
0 such that as |t| , wehave<br />
gt g0 gt 1/2. *<br />
1<br />
Let N be a positive integer so that<br />
2N<br />
Define two subsets U and V as follows:<br />
U <br />
V <br />
1<br />
, thus let ,0 : p and 1<br />
,0 : q.<br />
2N 2N1<br />
x, y : x p q<br />
2<br />
x, y : x p q<br />
2<br />
gq, p<br />
gq, p<br />
<strong>The</strong>n we have<br />
(1). U V gq, p, (2). U , sincep U and V , sinceq V,<br />
(3). U V by the given set A, and (*)<br />
Since x, y : x pq<br />
2<br />
gq, p. So, we have<br />
and x, y : x <br />
pq<br />
2 are open in R2 , then U and V are open in<br />
(4). U is open in gq, p and V is open in gq, p.<br />
From (1)-(4), we have gq, p is disconnected which is absurb since a connected subset<br />
under a continuous function is connected. So, such g cannot exist. It means that S is not<br />
arcwise connected.<br />
Remark: This exercise gives us an example to say that connectedness does not imply<br />
path-connectedness. <strong>And</strong> it is important example which is worth keeping in mind.<br />
4.47 Let F F 1 , F 2 ,... be a countable collection of connected compact sets in R n<br />
such that F k1 F k for each k 1. Prove that the intersection <br />
k1 F k is connected and<br />
closed.<br />
Proof: Since F k is compact for each k 1, F k is closed for each k 1. Hence,<br />
<br />
k1 F k : F is closed. Note that by <strong>The</strong>orem 3.39, we know that F is compact. Assume<br />
that F is not connected. <strong>The</strong>n there are two subsets A and B with<br />
1.A , B . 2.A B . 3.A B F. 4.A, B are closed in F.<br />
Note that A, B are closed and disjoint in R n . By exercise 4.57, there exist U and V which<br />
are open and disjoint in R n such that A U, andB V. Claim that there exists F k such<br />
that F k U V. Suppose NOT, thenthereexistsx k F k U V. Without loss of<br />
generality, we may assume that x k F k1 . So, we have a sequence x k F 1 which<br />
implies that there exists a convergent subsequence x kn , say lim kn x kn x. Itis<br />
clear that x F k for all k since x is an accumulation point of each F k . So, we have
x F <br />
k1 F k A B U V<br />
which implies that x is an interior point of U V since U and V are open. So,<br />
Bx; U V for some 0, which contradicts to the choice of x k . Hence, we have<br />
proved that there exists F k such that F k U V. LetC U F k ,andD V F k , then<br />
we have<br />
1. C since A U and A F k ,andD since B V and B F k .<br />
2. C D U F k V F k U V .<br />
3. C D U F k V F k F k .<br />
4. C is open in F k and D is open in F k by C, D are open in R n .<br />
Hence, we have F k is disconnected which is absurb. So, we know that F <br />
k1 F k is<br />
connected.<br />
4.48 Let S be an open connected set in R n .LetT be a component of R n S. Prove<br />
that R n T is connected.<br />
Proof: If S is empty, there is nothing to proved. Hence, we assume that S is nonempty.<br />
Write R n S xR n S Ux, whereUx is a component of R n S. So, we have<br />
R n S xR n S Ux.<br />
Say T Up, forsomep. <strong>The</strong>n<br />
R n T S xR n ST Ux.<br />
Claim that clS Ux for all x R n S T. If we can show the claim, given<br />
a, b R n T, and a two valued function on R n T. Note that clS is also connected. We<br />
consider three cases. (1) a S, b Ux for some x. (2)a, b S. (3)a Ux,<br />
b Ux .<br />
For case (1), let c clS Ux, then there are s n S and u n Ux with<br />
s n c and u n c, then we have<br />
fa lim n<br />
fs n f lim n<br />
s n fc f lim n<br />
u n lim n<br />
fu n fb<br />
which implies that fa fb.<br />
For case (2), it is clear fa fb since S itself is connected.<br />
For case (3), we choose s S, and thus use case (1), we know that<br />
fa fs fb.<br />
By case (1)-(3), we have f is constant on R n T. Thatis,R n T is connected.<br />
It remains to show the claim. To show clS Ux for all x R n S T, i.e., to<br />
show that for all x R n S T,<br />
clS Ux S S Ux<br />
S Ux<br />
.<br />
Suppose NOT, i.e., for some x, S Ux which implies that Ux R n clS<br />
which is open. So, there is a component V of R n clS contains Ux, whereV is open by<br />
<strong>The</strong>orem 4.44. However, R n clS R n S, sowehaveV is contained in Ux.<br />
<strong>The</strong>refore, we have Ux V. Note that Ux R n S, andR n S is closed. So,<br />
clUx R n S. By definition of component, we have clUx Ux, whichis<br />
closed. So, we have proved that Ux V is open and closed. It implies that Ux R n or<br />
which is absurb. Hence, the claim has proved.<br />
4.49 Let S, d be a connected metric space which is not bounded. Prove that for every
a in S and every r 0, the set x : dx, a r is nonempty.<br />
Proof: Assume that x : dx, a r is empty. Denote two sets x : dx, a r by A<br />
and x : dx, a r by B. <strong>The</strong>n we have<br />
1. A since a A and B since S is unbounded,<br />
2. A B ,<br />
3. A B S,<br />
4. A Ba; r is open in S,<br />
and consider B as follows. Since x : dx, a r is closed in S, B S x : dx, a r<br />
is open in S. So, we know that S is disconnected which is absurb. Hence, we know that the<br />
set x : dx, a r is nonempty.<br />
Supplement on a connected metric space<br />
Definition Two subsets A and B of a metric space X are said to be separated if both<br />
A clB and clA B .<br />
AsetE X is said to be connected if E is not a union of two nonempty separated<br />
sets.<br />
We now prove the definition of connected metric space is equivalent to this definiton<br />
as follows.<br />
<strong>The</strong>orem A set E in a metric space X is connected if, and only if E is not the union of<br />
two nonempty disjoint subsets, each of which is open in E.<br />
Proof: () Suppose that E is the union of two nonempty disjoint subsets, each<br />
of which is open in E, denote two sets, U and V. Claim that<br />
U clV clU V .<br />
Suppose NOT, i.e., x U clV. Thatis,thereisa 0 such that<br />
B X x, E B E x, U and B X x, V <br />
which implies that<br />
B X x, V B X x, V E<br />
B X x, E V<br />
U V ,<br />
a contradiction. So, we have U clV . Similarly for clU V . So,X is<br />
disconnected. That is, we have shown that if a set E in a metric space X is<br />
connected, then E is not the union of two nonempty disjoint subsets, each of which<br />
is open in E.<br />
() Suppose that E is disconnected, then E is a union of two nonempty<br />
separated sets, denoted E A B, whereA clB clA B . Claim that A<br />
and B are open in E. Suppose NOT, it means that there is a point x A which is<br />
not an interior point of A. So, for any ball B E x, r, there is a correspounding<br />
x r B, wherex r B E x, r. It implies that x clB which is absurb with<br />
A clB . So, we proved that A is open in E. Similarly, B is open in E. Hence,<br />
we have proved that if E is not the union of two nonempty disjoint subsets, each of<br />
whichisopeninE, then E in a metric space X is connected.<br />
Exercise Let A and B be connected sets in a metric space with A B not connected and<br />
suppose A B C 1 C 2 where clC 1 C 2 C 1 clC 2 . Show that<br />
B C 1 is connected.
Proof: Assume that B C 1 is disconnected, and thus we will prove that C 1 is<br />
disconnected. Consider, by clC 1 C 2 C 1 clC 2 ,<br />
C 1 clC 2 A B C 1 clA B C 1 clB *<br />
and<br />
clC 1 C 2 A B clC 1 A B clC 1 B **<br />
we know that at least one of (*) and (**) is nonempty by the hypothesis A is<br />
connected. In addition, by (*) and (**), we know that at leaset one of<br />
C 1 clB<br />
and<br />
clC 1 B<br />
is nonempty. So, we know that C 1 is disconnected by the hypothesis B is connected,<br />
and the concept of two valued function.<br />
From above sayings and hypothesis, we now have<br />
1. B is connected.<br />
2. C 1 is disconnected.<br />
3. B C 1 is disconnected.<br />
Let D be a component of B C 1 so that B D; we have, let<br />
B C 1 D E C 1 ,<br />
D clE clD E <br />
which implies that<br />
clE A E , andclA E E .<br />
So,wehaveprovethatA is disconnected wich is absurb. Hence, we know that<br />
B C 1 is connected.<br />
Remark We prove that clA E E clE A E as follows.<br />
Proof: Since<br />
D clE ,<br />
we obtain that<br />
clE A E<br />
clE D C 2 A B<br />
clE D C 2 B<br />
clE D C 2 since B D<br />
clE C 2 since D clE <br />
<strong>And</strong> since<br />
we obtain that<br />
clC 1 C 2 since E C 1<br />
.<br />
clD E ,
clA E E<br />
clD C 2 A B E<br />
clD C 2 B E<br />
clD C 2 E since B D<br />
clC 2 E since clD E <br />
clC 2 C 1 since E C 1<br />
.<br />
Exercise Prove that every connected metric space with at least two points is uncountable.<br />
Proof: LetX be a connected metric space with two points a and b, where<br />
a b. Define a set A r x : dx, a r and B r x : dx, a r. It is clear<br />
that both of sets are open and disjoint. Assume X is countable. Let<br />
da,b<br />
r , da,b , it guarantee that both of sets are non-empty. Since<br />
4 2<br />
da,b<br />
, da,b is uncountable, we know that there is a 0 such that<br />
4 2<br />
A B X. It implies that X is disconnected. So, we know that such X is<br />
countable.<br />
Uniform continuity<br />
4.50 Prove that a function which is uniformly continuous on S is also continuous on S.<br />
Proof: Let f be uniformly continuous on S, thengiven 0, there exists a 0 such<br />
that as dx, y , x and y in S, then we have<br />
dfx, fy .<br />
Fix y, called a. <strong>The</strong>ngiven 0, there exists a 0 such that as dx, a , x in S,<br />
then we have<br />
dfx, fa .<br />
That is, f is continuous at a. Sincea is arbitrary, we know that f is continuous on S.<br />
4.51 If fx x 2 for x in R, prove that f is not uniformly continuous on R.<br />
Proof: Assume that f is uniformly continuous on R, thengiven 1, there exists a<br />
0 such that as |x y| , wehave<br />
|fx fy| 1.<br />
Choose x y ,(<br />
2<br />
|x y| , then we have<br />
2<br />
|fx fy| y 1. 2<br />
When we choose y 1 , then 1 <br />
2<br />
2<br />
1 <br />
2<br />
1<br />
which is absurb. Hence, we know that f is not uniformly continuous on R.<br />
Remark: <strong>The</strong>re are some similar questions written below.<br />
1. Here is a useful lemma to make sure that a function is uniformly continuous on<br />
a, b, but we need its differentiability.<br />
(Lemma) Letf : a, b R R be differentiable and |f x| M for all x a, b.<br />
<strong>The</strong>n f is uniformly continuous on a, b, wherea, b may be .
Proof: By Mean Value <strong>The</strong>orem, wehave<br />
|fx fy| |f z||x y|, wherez x, y or y, x<br />
M|x y| by hypothesis. *<br />
<strong>The</strong>n given 0, there is a /M such that as |x y| , x, y a, b, wehave<br />
|fx fy| , by (*).<br />
Hence, we know that f is uniformly continuous on a, b.<br />
Note: A standard example is written in Remark 2. But in Remark 2, we still use<br />
definition of uniform continuity to practice what it says.<br />
2. sin x is uniformly continuous on R.<br />
Proof: Given 0, we want to find a 0 such that as |x y| , wehave<br />
Since sin x sin y 2cos xy<br />
2<br />
|sin x sin y| .<br />
xy<br />
sin ,<br />
2<br />
|sin x| |x|, and|cosx| 1, we have<br />
|sin x sin y| |x y|<br />
So, if we choose , then as |x y| , it implies that<br />
|sin x sin y| .<br />
That is, sinx is uniformly continuous on R.<br />
Note: |sin x sin y| |x y| for all x, y R, can be proved by Mean Value <strong>The</strong>orem<br />
as follows.<br />
proof: By Mean Value <strong>The</strong>orem, sin x sin y sin z x y; itimpliesthat<br />
|sin x sin y| |x y|.<br />
3. sinx 2 is NOT uniformly continuous on R.<br />
Proof: Assume that sinx 2 is uniformly continuous on R. <strong>The</strong>ngiven 1, there is a<br />
0 such that as |x y| , wehave<br />
|sinx 2 siny 2 | 1. *<br />
Consider<br />
n <br />
2 n 2<br />
n n 0,<br />
4 n<br />
2<br />
and thus choose N <br />
<br />
<br />
1 <br />
4 2 4 2<br />
N 2<br />
which implies<br />
N .<br />
So, choose x N 2<br />
and y N , thenby(*),wehave<br />
|fx fy| sin N sinN sin<br />
2<br />
1 1<br />
2<br />
which is absurb. So, sinx 2 is not uniformly continuous on R.<br />
4. x is uniformly continuous on 0, .<br />
Proof: Since x y |x y| for all x, y 0, , thengiven 0, there exists<br />
a 2 such that as |x y| , x, y 0, , wehave<br />
So, we know that<br />
x y |x y| .<br />
x is uniformly continuous on 0, .
Note: We have the following interesting results:. Prove that, for x 0, y 0,<br />
|x p y p | <br />
|x y| p if 0 p 1,<br />
p|x y|x p1 y p1 if 1 p .<br />
Proof: (As 0 p 1) Without loss of generality, let x y, consider<br />
fx x y p x p y p , then<br />
f x p x y p1 x p1 0, note that p 1 0.<br />
So, we have f is an increasing function defined on 0, for all given y 0. Hence, we<br />
have fx f0 0. So,<br />
x p y p x y p if x y 0<br />
which implies that<br />
|x p y p | |x y| p<br />
for x 0, y 0.<br />
Ps: <strong>The</strong> inequality, we can prove the case p 1/2 directly. Thus the inequality is not<br />
surprising for us.<br />
(As 1 p Without loss of generality, let x y, consider<br />
x p y p pz p1 x y, wherez y, x, byMean Value <strong>The</strong>orem.<br />
which implies<br />
for x 0, y 0.<br />
px p1 x y, note that p 1 0,<br />
px p1 y p1 x y<br />
|x p y p | p|x y|x p1 y p1 <br />
5. In general, we have<br />
x r is uniformly continuous on 0, , ifr 0, 1,<br />
is NOT uniformly continuous on 0, , ifr 1,<br />
and<br />
sinx r <br />
is uniformly continuous on 0, , ifr 0, 1,<br />
is NOT uniformly continuous on 0, , ifr 1.<br />
Proof: (x r )Asr 0, it means that x r is a constant function. So, it is obviuos. As<br />
r 0, 1, thengiven 0, there is a 1/r 0 such that as |x y| , x, y 0, ,<br />
we have<br />
|x r y r | |x y| r by note in the exercise<br />
r<br />
.<br />
So, x r is uniformly continuous on 0, , ifr 0, 1.<br />
As r 1, assume that x r is uniformly continuous on 0, , thengiven 1 0,<br />
there exists a 0 such that as |x y| , x, y 0, , wehave<br />
|x r y r | 1. *<br />
By Mean Value <strong>The</strong>orem, we have (let x y /2, y 0)
x r y r rz r1 x y<br />
ry r1 /2.<br />
So, if we choose y 2 1<br />
r1<br />
, then we have<br />
r<br />
x r y r 1<br />
which is absurb with (*). Hence, x r is not uniformly continuous on 0, .<br />
Ps: <strong>The</strong> reader should try to realize why x r is not uniformly continuous on 0, , for<br />
r 1. <strong>The</strong> ruin of non-uniform continuity comes from that x is large enough. At the same<br />
time, compare it with theorem that a continuous function defined on a compact set K is<br />
uniflormly continuous on K.<br />
(sin x r )Asr 0, it means that x r is a constant function. So, it is obviuos. As r 0, 1,<br />
given 0, there is a 1/r 0 such that as |x y| , x, y 0, , wehave<br />
|sin x r sin y r x<br />
| 2cos<br />
r y r<br />
sin<br />
xr y r<br />
2<br />
2<br />
|x r y r |<br />
|x y| r by the note in the Remark 4.<br />
r<br />
.<br />
So, sin x r is uniformly continuous on 0, , ifr 0, 1.<br />
As r 1, assume that sin x r is uniformly continuous on 0, , thengiven 1, there<br />
is a 0 such that as |x y| , x, y 0, , wehave<br />
|sin x r sin y r | 1. **<br />
Consider a sequence n 2 1/r n 1/r , it is easy to show that the sequence tends to<br />
0asn . So, there exists a positive integer N such that |x y| , x n 2 1/r ,<br />
y n 1/r . <strong>The</strong>n<br />
sin x r sin y r 1<br />
which contradicts (**). So, we know that sin x r is not uniformly continuous on 0, .<br />
Ps: For n 2 1/r n 1/r : x n 0asn 0, here is a short proof by using<br />
L-Hospital Rule.<br />
Proof: Write<br />
x n n 1/r<br />
n<br />
1/r<br />
2<br />
n 1/r 1 1 2n<br />
1/r<br />
1<br />
and thus consider the following limit<br />
<br />
1 1<br />
2n 1/r 1<br />
n 1/r
lim x<br />
1 1 2x 1/r 1<br />
x 1/r , 0<br />
0<br />
lim 1/r<br />
x 2 x 1 r 1<br />
1 <br />
2x<br />
1 1 r 1<br />
by L-Hospital Rule.<br />
0.<br />
Hence x n 0asn .<br />
6. Here is a useful criterion for a function which is NOT uniformly continuous defined<br />
a subset A in a metric space. We say a function f is not uniformly continuous on a subset A<br />
in a metric space if, and only if, there exists 0 0, and two sequences x n and y n <br />
such that as<br />
lim n<br />
x n y n 0<br />
which implies that<br />
|fx n fy n | 0 for n is large enough.<br />
<strong>The</strong> criterion is directly from the definition on uniform continuity. So, we omit the<br />
proof.<br />
4.52 Assume that f is uniformly continuous on a bounded set S in R n . Prove that f<br />
must be bounded on S.<br />
Proof: Since f is uniformly continuous on a bounded set S in R n ,given 1, then<br />
there exists a 0 such that as x y , x, y S, wehave<br />
dfx, fy 1.<br />
Consider the closure of S, clS is closed and bounded. Hence clS is compact. <strong>The</strong>n for<br />
any open covering of clS, there is a finite subcover. That is,<br />
clS xclS Bx; /2,<br />
clS kn k1 Bx k ; /2, wherex k clS,<br />
S kn k1 Bx k ; /2, wherex k clS.<br />
Note that if Bx k ; /2 S for some k, then we remove this ball. So, we choose<br />
y k Bx k ; /2 S, 1 k n and thus we have<br />
Bx k ; /2 By k ; for 1 k n,<br />
since let z Bx k ; /2,<br />
z y k z x k x k y k /2 /2 .<br />
Hence, we have<br />
S kn k1 By k ; , wherey k S.<br />
Given x S, then there exists By k ; for some k such that x By k ; . So,<br />
dfx, fx k 1 fx Bfy k ;1<br />
Note that kn k1 Bfy k ;1 is bounded since every Bfy k ;1 is bounded. So, let B be a<br />
bounded ball so that kn k1 Bfy k ;1 B. Hence, we have every x S, fx B. Thatis,<br />
f is bounded.<br />
Remark: If we know that the codomain is complete, then we can reduce the above<br />
proof. See Exercise 4.55.<br />
4.53 Let f be a function defined on a set S in R n and assume that fS R m .Letg be<br />
defined on fS with value in R k , and let h denote the composite function defined by
hx gfx if x S. Iff is uniformly continuous on S and if g is uniformly continuous<br />
on fS, show that h is uniformly continuous on S.<br />
Proof: Given 0, we want to find a 0 such that as x y R<br />
n , x, y S, we<br />
have<br />
hx hy gfx gfy .<br />
For the same , sinceg is uniformly continuous on fS, then there exists a 0 such<br />
that as fx fy R<br />
m ,wehave<br />
gfx gfy .<br />
For this ,sincef is uniformly continuous on S, then there exists a 0 such that as<br />
x y R<br />
n , x, y S, wehave<br />
fx fy R<br />
m .<br />
So, given 0, there is a 0 such that as x y R<br />
n , x, y S, wehave<br />
hx hy .<br />
That is, h is uniformly continuous on S.<br />
Remark: It should be noted that (Assume that all functions written are continuous)<br />
(1) uniform continuity uniform continuity uniform continuity.<br />
(2) uniform continuity NOT uniform continuity <br />
(3) NOT uniform continuity uniform continuity <br />
(a) NOT uniform continuity, or<br />
(b) uniform continuity.<br />
(a) NOT uniform continuity, or<br />
(b) uniform continuity.<br />
(4) NOT uniform continuity NOT uniform continuity <br />
(a) NOT uniform continuity, or<br />
(b) uniform continuity.<br />
For (1), it is from the exercise.<br />
For (2), (a) let fx x, andgx x 2 , x R fgx fx 2 x 2 .<br />
(b) let fx x ,andgx x 2 , x 0, fgx fx 2 x.<br />
For (3), (a) let fx x 2 ,andgx x, x R fgx fx x 2 .<br />
(b) let fx x 2 ,andgx x , x 0, fgx f x x.<br />
For (4), (a) let fx x 2 ,andgx x 3 , x R fgx fx 3 x 6 .<br />
(b) let fx 1/x, andgx 1 , x 0, 1 fgx f 1<br />
x .<br />
x x<br />
Note. In (4), we have x r is not uniformly continuous on 0, 1, forr 0. Here is a<br />
proof.<br />
Proof: Let r 0, and assume that x r is not uniformly continuous on 0, 1. Given<br />
1, there is a 0 such that as |x y| , wehave<br />
|x r y r | 1. *<br />
Let x n 2/n, andy n 1/n. <strong>The</strong>n x n y n 1/n. Choose n large enough so that 1/n .<br />
So, we have
|x r y r | 2 n<br />
r<br />
1 n<br />
r<br />
r<br />
1 n |2r 1| , asn since r 0,<br />
which is absurb with (*). Hence, we know that x r is not uniformly continuous on 0, 1, for<br />
r 0.<br />
Ps: <strong>The</strong> reader should try to realize why x r is not uniformly continuous on 0, 1, for<br />
r 0. <strong>The</strong> ruin of non-uniform continuity comes from that x is small enough.<br />
4.54 Assume f : S T is uniformly continuous on S, whereS and T are metric<br />
spaces. If x n is any Cauchy sequence in S, prove that fx n is a Cauchy sequence in T.<br />
(Compare with Exercise 4.33.)<br />
Proof: Given 0, we want to find a positive integer N such that as n, m N, we<br />
have<br />
dfx n , fx m .<br />
For the same , sincef is uniformly continuous on S, then there is a 0 such that as<br />
dx, y , x, y S, wehave<br />
dfx, fy .<br />
For this , sincex n is a Cauchy sequence in S, then there is a positive integer N such<br />
that as n, m N, wehave<br />
dx n , x m .<br />
Hence, given 0, there is a postive integer N such that as n, m N, wehave<br />
dfx n , fx m .<br />
That is, fx n is a Cauchy sequence in T.<br />
Remark: <strong>The</strong> reader should compare with Exercise 4.33 and Exercise 4.55.<br />
4.55 Let f : S T be a function from a metric space S to another metric space T.<br />
Assume that f is uniformly continuous on a subset A of S and let T is complete. Prove that<br />
there is a unique extension of f to clA which is uniformly continuous on clA.<br />
Proof: Since clA A A , it suffices to consider the case x A A. Since<br />
x A A, then there is a sequence x n A with x n x. Note that this sequence is a<br />
Cauchy sequence, so we have by Exercise 4.54, fx n is a Cauchy sequence in T since f<br />
is uniformly on A. In addition, since T is complete, we know that fx n is a convergent<br />
sequence, say its limit L. Note that if there is another sequence x n A with x n x,<br />
then fx n is also a convergent sequence, say its limit L . Note that x n x n is still a<br />
Cauchy sequence. So, we have<br />
dL, L dL, fx n dfx n , fx n dfx n , L 0asn .<br />
So, L L . That is, it is well-defined for g : clA T by the following<br />
gx <br />
fx if x A,<br />
lim n fx n if x A A, wherex n x.<br />
So, the function g is a extension of f to clA.<br />
Claim that this g is uniformly continuous on clA. That is, given 0, we want to<br />
find a 0 such that as dx, y , x, y clA, wehave<br />
dgx, gy .<br />
Since f is uniformly continuous on A, for /3, there is a 0 such that as
dx, y , x, y A, wehave<br />
dfx, fy .<br />
Let x, y clA, and thus we have x n A with x n x, andy n A with y n y.<br />
Choose /3, then we have<br />
dx n , x /3 and dy n , y /3 as n N 1<br />
So, as dx, y /3, wehave(n N 1 )<br />
dx n , y n dx n , x dx, y dy, y n /3 /3 /3 .<br />
Hence, we have as dx, y , (n N 1 )<br />
dgx, gy dgx, fx n dfx n , fy n dfy n , fy<br />
*<br />
dgx, fx n dfy n , gy<br />
<strong>And</strong> since lim n fx n gx, and lim n fy n gy, we can choose N N 1 such that<br />
dgx, fx n and<br />
dfy n , gy .<br />
So, as dx, y , (n N) wehave<br />
dgx, gy 3 by (*).<br />
That is, g is uniformly on clA.<br />
It remains to show that g is a unique extension of f to clA which is uniformly<br />
continuous on clA. If there is another extension h of f to clA which is uniformly<br />
continuous on clA, thengivenx A A, we have, by continuity, (Say x n x)<br />
hx h lim n<br />
x n lim n<br />
hx n lim n<br />
fx n lim n<br />
gx n g lim n<br />
x n gx<br />
which implies that hx gx for all x A A. Hence, we have hx gx for all<br />
x clA. Thatis,g is a unique extension of f to clA which is uniformly continuous on<br />
clA.<br />
Remark: 1. We do not require that A is bounded, in fact, A is any non-empty set in a<br />
metric space.<br />
2. <strong>The</strong> exercise is a criterion for us to check that a given function is NOT uniformly<br />
continuous. For example, let f : 0, 1 R by fx 1/x. Sincef0 does not exist, we<br />
know that f is not uniformly continuous. <strong>The</strong> reader should feel that a uniformly continuous<br />
is sometimes regarded as a smooth function. So, it is not surprising for us to know the<br />
exercise. Similarly to check fx x 2 , x R, and so on.<br />
3. Here is an exercise to make us know that a uniformly continuous is a smooth<br />
function. Let f : R R be uniformly continuous, then there exist , 0 such that<br />
|fx| |x| .<br />
Proof: Since f is uniformly continuous on R, given 1, there is a 0 such that as<br />
|x y| , wehave<br />
|fx fy| 1. *<br />
Given any x R, then there is the positive integer N such that N |x| N 1. If<br />
x 0, we consider<br />
y 0 0, y 1 /2, y 2 ,...,y 2N1 N 2 , y 2N x.<br />
<strong>The</strong>n we have
N<br />
|fx f0| |fy 2k fy 2k1 | |fy 2k1 fy 2k2 |<br />
k1<br />
2N by (*)<br />
which implies that<br />
|fx| 2N |f0|<br />
2 1 |x| |f0| since |x| N 1<br />
<br />
2 |x| 2 |f0|.<br />
<br />
Similarly for x 0. So, we have proved that |fx| |x| for all x.<br />
4.56 In a metric space S, d, letA be a nonempty subset of S. Define a function<br />
f A : S R by the equation<br />
f A x infdx, y : y A<br />
for each x in S. <strong>The</strong> number f A x is called the distance from x to A.<br />
(a) Prove that f A is uniformly continuous on S.<br />
(b) Prove that clA x : x S and f A x 0 .<br />
Proof: (a) Given 0, we want to find a 0 such that as dx 1 , x 2 , x 1 , x 2 S,<br />
we have<br />
|f A x 1 f A x 2 | .<br />
Consider (x 1 , x 2 , y S)<br />
dx 1 , y dx 1 , x 2 dx 2 , y, anddx 2 , y dx 1 , x 2 dx 1 , y<br />
So,<br />
infdx 1 , y : y A dx 1 , x 2 infdx 2 , y : y A and<br />
infdx 2 , y : y A dx 1 , x 2 infdx 1 , y : y A<br />
which implies that<br />
f A x 1 f A x 2 dx 1 , x 2 and f A x 2 f A x 1 dx 1 , x 2 <br />
which implies that<br />
|f A x 1 f A x 2 | dx 1 , x 2 .<br />
Hence, if we choose , thenwehaveasdx 1 , x 2 , x 1 , x 2 S, wehave<br />
|f A x 1 f A x 2 | .<br />
That is, f A is uniformly continuous on S.<br />
(b) Define K x : x S and f A x 0 , we want to show clA K. We prove it<br />
by two steps.<br />
() Letx clA, then Bx; r A for all r 0. Choose y k Bx;1/k A, then<br />
we have<br />
infdx, y : y A dx, y k 0ask .<br />
So, we have f A x infdx, y : y A 0. So, clA K.<br />
() Letx K, then f A x infdx, y : y A 0. That is, given any 0, there<br />
is an element y A such that dx, y . Thatis,y Bx; A. So,x is an adherent<br />
point of A. Thatis,x clA. So, we have K clA.<br />
From above saying, we know that clA x : x S and f A x 0 .<br />
Remark: 1. <strong>The</strong> function f A often appears in Analysis, so it is worth keeping it in mind.
In addition, part (b) comes from intuition. <strong>The</strong> reader may think it twice about distance 0.<br />
2. Here is a good exercise to pratice. <strong>The</strong> statement is that suppose that K and F are<br />
disjoint subsets in a metric space X, K is compact, F is closed. Prove that there exists a<br />
0 such that dp, q if p K, q F. Show that the conclusion is may fail for two<br />
disjoint closed sets if neither is compact.<br />
Proof: Suppose NOT, i.e., for any 0, there exist p K, andq F such that<br />
dp , q . Let 1/n, then there exist two sequence p n K, andq n F such<br />
that dp n , q n 1/n. Note that p n K, andK is compact, then there exists a<br />
subsequence p nk with lim nk p nk p K. Hence, we consider dp nk , q nk n 1 k<br />
to get a<br />
contradiction. Since<br />
dp nk , p dp, q nk dp nk , q nk n 1 , k<br />
then let n k , wehavelim nk q nk p. Thatis,p is an accumulation point of F which<br />
implies that p F. So, we get a contradiction since K F . That is, there exists a<br />
0 such that dp, q if p K, q F.<br />
We give an example to show that the conclusion does not hold. Let<br />
K x,0 : x R and F x,1/x : x 0, then K and F are closd. It is clear that<br />
such cannot be found.<br />
Note: Two disjoint closed sets may has the distance 0, however; if one of closed sets is<br />
compact, then we have a distance 0. <strong>The</strong> reader can think of them in R n , and note that<br />
a bounded and closed subsets in R n is compact. It is why the example is given.<br />
4.57 In a metric space S, d, letA and B be disjoint closed subsets of S. Prove that<br />
there exists disjoint open subsets U and V of S such that A U and B V. Hint. Let<br />
gx f A x f B x, in the notation of Exercise 4.56, and consider g 1 ,0 and<br />
g 1 0, .<br />
Proof: Let gx f A x f B x, then by Exercixe 4.56, we have gx is uniformly<br />
continuous on S. So,gx is continuous on S. Consider g 1 ,0 and g 1 0, , and<br />
note that A, B are disjoint and closed, then we have by part (b) in Exercise 4.56,<br />
gx 0ifx A and<br />
gx 0ifx B.<br />
So, we have A g 1 ,0 : U, andB g 1 0, : V.<br />
Discontinuities<br />
4.58 Locate and classify the discontinuities of the functions f defined on R 1 by the<br />
following equations:<br />
(a) fx sin x/x if x 0, f0 0.<br />
sinx<br />
Solution: f is continuous on R 0, and since lim x0 x 1, we know that f has a<br />
removable discontinuity at 0.<br />
(b) fx e 1/x if x 0, f0 0.<br />
Solution: f is continuous on R 0, and since lim x0<br />
e 1/x and lim x0<br />
e 1/x 0,<br />
we know that f has an irremovable discontinuity at 0.<br />
(c) fx e 1/x sin 1/x if x 0, f0 0.<br />
Solution: f is continuous on R 0, and since the limit fx does not exist as x 0,<br />
we know that f has a irremovable discontinuity at 0.
(d) fx 1/1 e 1/x if x 0, f0 0.<br />
Solution: f is continuous on R 0, and since lim x0<br />
e 1/x and lim x0<br />
e 1/x 0,<br />
we know that f has an irremovabel discontinuity at 0. In addition, f0 0and<br />
f0 1, we know that f has the lefthand jump at 0, f0 f0 1, and f is<br />
continuous from the right at 0.<br />
4.59 Locate the points in R 2 at which each of the functions in Exercise 4.11 is not<br />
continuous.<br />
(a) By Exercise 4.11, we know that fx, y is discontinuous at 0, 0, where<br />
fx, y x2 y 2<br />
if x, y 0, 0, andf0, 0 0.<br />
x 2 y2 Let gx, y x 2 y 2 ,andhx, y x 2 y 2 both defined on R 2 0, 0, we know that g<br />
and h are continuous on R 2 0, 0. Note that h 0onR 2 0, 0. Hence, f g/h is<br />
continuous on R 2 0, 0.<br />
(b) By Exercise 4.11, we know that fx, y is discontinuous at 0, 0, where<br />
xy 2<br />
fx, y <br />
if x, y 0, 0, andf0, 0 0.<br />
xy 2 2<br />
x y<br />
Let gx, y xy 2 ,andhx, y xy 2 x y 2 both defined on R 2 0, 0, we know<br />
that g and h are continuous on R 2 0, 0. Note that h 0onR 2 0, 0. Hence,<br />
f g/h is continuous on R 2 0, 0.<br />
(c) By Exercise 4.11, we know that fx, y is continuous at 0, 0, where<br />
fx, y 1 x sinxy if x 0, and f0, y y,<br />
since lim x,y0,0 fx, y 0 f0, 0. Letgx, y 1/x and hx, y sinxy both defined<br />
on R 2 0, 0, we know that g and h are continuous on R 2 0, 0. Note that h 0<br />
on R 2 0, 0. Hence, f g/h is continuous on R 2 0, 0. Hence, f is continuous on<br />
R 2 .<br />
(d) By Exercise 4.11, we know that fx, y is continuous at 0, 0, where<br />
x y sin1/x sin1/y if x 0andy 0,<br />
fx, y <br />
0 if x 0ory 0.<br />
since lim x,y0,0 fx, y 0 f0, 0. It is the same method as in Exercise 4.11, we know<br />
that f is discontinuous at x,0 for x 0andf is discontinuous at 0, y for y 0. <strong>And</strong> it is<br />
clearly that f is continuous at x, y, wherex 0andy 0.<br />
(e) By Exercise 4.11, Since<br />
we rewrite<br />
fx, y <br />
fx, y <br />
cos<br />
sinxsiny<br />
tan xtan y<br />
,iftanx tan y,<br />
cos 3 x if tan x tan y.<br />
xy<br />
cos xcos y<br />
2<br />
cos<br />
xy<br />
2<br />
if tan x tan y<br />
cos 3 x if tan x tan y.<br />
We consider x, y /2, /2 /2, /2, others are similar. Consider two cases (1)<br />
x y, and (2) x y, wehave<br />
(1) (x y) Since lim x,ya,a fx, y cos 3 a fa, a. Hence, we know that f is<br />
,<br />
.
continuous at a, a.<br />
(2) (x y) Sincex y, itimpliesthattanx tan y. Note that the denominator is not 0<br />
since x, y /2, /2 /2, /2. So, we know that f is continuous at a, b, a b.<br />
So, we know that f is continuous on /2, /2 /2, /2.<br />
Monotonic functions<br />
4.60 Let f be defined in the open interval a, b and assume that for each interior point x<br />
of a, b there exists a 1 ball Bx in which f is increasing. Prove that f is an increasing<br />
function throughout a, b.<br />
Proof: Suppose NOT, i.e., there exist p, q with p q such that fp fq. Consider<br />
p, q a, b, and since for each interior point x of a, b there exists a 1 ball Bx in<br />
which f is increasing. <strong>The</strong>n p, q xp,q Bx; x , (<strong>The</strong> choice of balls comes from the<br />
hypothesis). It implies that p, q n<br />
k1 Bx n ; n : B n . Note that if B i B j , we remove<br />
such B i and make one left. Without loss of generality, we assume that x 1 .. x n .<br />
.fp fx 1 ... fx n fq<br />
which is absurb. So, we know that f is an increasing function throughout a, b.<br />
4.61 Let f be continuous on a compact interval a, b and assume that f does not have a<br />
loacal maximum or a local minimum at any interior point. (See the note following Exercise<br />
4.25.) Prove that f must be monotonic on a, b.<br />
Proof: Since f is continuous on a, b, wehave<br />
max fx fp, wherep a, b and<br />
xa,b<br />
min fx fq, whereq a, b.<br />
xa,b<br />
So, we have p, q a, b by hypothesis that f does not have a local maximum or a local<br />
minimum at any interior point. Without loss of generality, we assume that p a, and<br />
q b. Claim that f is decreasing on a, b as follows.<br />
Suppose NOT, then there exist x, y a, b with x y such that fx fy. Consider<br />
x, y and by hyothesis, we know that f| x,y has the maximum at y, andf| a,y has the<br />
minimum at y. <strong>The</strong>n it implies that there exists By; x, y such that f is constant on<br />
By; x, y, which contradicts to the hypothesis. Hence, we have proved that f is<br />
decreasing on a, b.<br />
4.62 If f is one-to one and continuous on a, b, prove that f must be strictly<br />
monotonic on a, b. That is, prove that every topological mapping of a, b onto an<br />
interval c, d must be strictly monotonic.<br />
Proof: Since f is continuous on a, b, wehave<br />
max fx fp, wherep a, b and<br />
xa,b<br />
min fx fq, whereq a, b.<br />
xa,b<br />
Assume that p a, b, then there exists a 0 such that fy fp for all<br />
y p , p a, b. Choose y 1 x , x and y 2 x, x , thenwehaveby<br />
1-1, fy 1 fx and fy 2 fx. <strong>And</strong> thus choose r so that<br />
fy 1 r fx fz 1 r, wherez 1 y 1 , x by Intermediate Value <strong>The</strong>orem,<br />
fy 2 r fx fz 2 r, wherez 2 x, y 2 by Intermediate Value <strong>The</strong>orem,<br />
which contradicts to 1-1. So, we know that p a, b. Similarly, we have q a, b.
Without loss of generality, we assume that p a and q b. Claim that f is strictly<br />
decreasing on a, b.<br />
Suppose NOT, then there exist x, y a, b, with x y such that fx fy. (""<br />
does not hold since f is 1-1.) Consider x, y and by above method, we know that f| x,y has<br />
the maximum at y, andf| a,y has the minimum at y. <strong>The</strong>n it implies that there exists<br />
By; x, y such that f is constant on By; x, y, which contradicts to 1-1. Hence,<br />
for any x y a, b, wehavefx fy. ("" does not hold since f is 1-1.) So, we<br />
have proved that f is strictly decreasing on a, b.<br />
Reamrk: 1. Here is another proof by Exercise 4.61. It suffices to show that 1-1 and<br />
continuity imply that f does not have a local maximum or a local minimum at any interior<br />
point.<br />
Proof: Suppose NOT, it means that f has a local extremum at some interior point x.<br />
Without loss of generality, we assume that f has a local minimum at the interior point x.<br />
Since x is an interior point of a, b, then there exists an open interval<br />
x , x a, b such that fy fx for all y x , x . Note that f is 1-1, so<br />
we have fy fx for all y x , x x. Choose y 1 x and<br />
y 2 x, x , then we have fy 1 fx and fy 2 fx. <strong>And</strong> thus choose r so that<br />
fy 1 r fx fp r, wherep y 1 , x by Intermediate Value <strong>The</strong>orem,<br />
fy 2 r fx fq r, whereq x, y 2 by Intermediate Value <strong>The</strong>orem,<br />
which contradicts to the hypothesis that f is 1-1. Hence, we have proved that 1-1 and<br />
continuity imply that f does not have a local maximum or a local minimum at any interior<br />
point.<br />
2. Under the assumption of continuity on a compact interval, one-to-one is<br />
equivalent to being strictly monotonic.<br />
Proof: By the exercise, we know that an one-to-one and continuous function defined<br />
on a compact interval implies that a strictly monotonic function. So, it remains to show that<br />
a strictly monotonic function implies that an one-to-one function. Without loss of<br />
generality, let f be increasing on a, b, then as fx fy, we must have x y since if<br />
x y, then fx fy and if x y, then fx fy. So, we have proved that a strictly<br />
monotonic function implies that an one-to-one function. Hence, we get that under the<br />
assumption of continuity on a compact interval, one-to-one is equivalent to being strictly<br />
monotonic.<br />
4.63 Let f be an increasing function defined on a, b and let x 1 ,..,x n be n points in<br />
the interior such that a x 1 x 2 ... x n b.<br />
n<br />
(a) Show that k1<br />
fx k fx k fb fa .<br />
Proof: Let a x 0 and b x n1 ;sincef is an increasing function defined on a, b, we<br />
know that both fx k and fx k exist for 1 k n. Assume that y k x k , x k1 , then<br />
we have fy k fx k and fx k1 fy k1 . Hence,<br />
n<br />
<br />
k1<br />
fx k fx k fy k fy k1 <br />
n<br />
k1<br />
fy n fy 0 <br />
fb fa .<br />
(b) Deduce from part (a) that the set of dicontinuities of f is countable.
Proof: Let D denote the set of dicontinuities of f. Consider<br />
D m x a, b : fx fx m 1 , then D <br />
m1 D m . Note that #D m , so<br />
we have D is countable. That is, the set of dicontinuities of f is countable.<br />
(c) Prove that f has points of continuity in every open subintervals of a, b.<br />
Proof: By (b), f has points of continuity in every open subintervals of a, b, since<br />
every open subinterval is uncountable.<br />
Remark: (1) Here is another proof about (b). Denote Q x 1 ,...,x n ,..., and let x be<br />
a point at which f is not continuous. <strong>The</strong>n we have fx fx 0. (If x is the end<br />
point, we consider fx fx 0orfx fx 0) So, we have an open interval I x<br />
such that I x fa, b fx. <strong>The</strong> interval I x contains infinite many rational numbers,<br />
we choose the smallest index, say m mx. <strong>The</strong>n the number of the set of discontinuities<br />
of f on a, b is a subset of N. Hence, the number of the set of discontinuities of f on a, b<br />
is countable.<br />
(2) <strong>The</strong>re is a similar exercise; we write it as a reference. Let f be a real valued function<br />
defined on 0, 1. Suppose that there is a positive number M having the following<br />
condition: for every choice of a finite number of points x 1 ,..,x n in 0, 1, wehave<br />
n<br />
M i1<br />
x i M. Prove that S : x 0, 1 : fx 0 is countable.<br />
Proof: Consider S n x 0, 1 : |fx| 1/n, then it is clear that every S n is<br />
countable. Since S <br />
n1 S n , we know that S is countable.<br />
4.64 Give an example of a function f, defined and strictly increasing on a set S in R,<br />
such that f 1 is not continuous on fS.<br />
Solution: Let<br />
x if x 0, 1,<br />
fx <br />
1ifx 2.<br />
<strong>The</strong>n it is clear that f is strictly increasing on 0, 1, sof has the incerse function<br />
x if x 0, 1,<br />
f 1 x <br />
2ifx 1.<br />
which is not continuous on fS 0, 1.<br />
Remark: Compare with Exercise 4.65.<br />
4.65 Let f be strictly increasing on a subset S of R. Assume that the image fS has one<br />
of the following properties: (a) fS is open; (b) fS is connected; (c) fS is closed. Prove<br />
that f must be continuous on S.<br />
Proof: (a) Given a S, then fa fS. Given 0, wewantofinda 0 such<br />
that as x Ba; S, wehave|fx fa| . SincefS is open, then there exists<br />
Bfa, fS, where .<br />
Claim that there exists a 0 such that fBa; S Bfa, . Choose<br />
y 1 fa /2 and y 2 fa /2, then y 1 fx 1 and y 2 fx 2 ,wehave<br />
x 1 a x 2 since f is strictly increasing on S. Hence, for x x 1 , x 2 S, wehave<br />
fx 1 fx fx 2 since f is strictly increasing on S. So,fx Bfa, .Let<br />
mina x 1 , x 2 a, then Ba; S a , a S x 1 , x 2 S which implis<br />
that fBa; S Bfa, .( Bfa, )<br />
Hence we have prove the claim, and the claim tells us that f is continuous at a. Sincea<br />
is arbitrary, we know that f is continuous on S.
(b) Note that since fS R, andfS is connected, we know that fS is an interval I.<br />
Given a S, then fa I. We discuss 2 cases as follows. (1) fa is an interior point of I.<br />
(2) fa is the endpoint of I.<br />
For case (1), it is similar to (a). We omit the proof.<br />
For case (2), it is similar to (a). We omit the proof.<br />
So, we have proved that f is continuous on S.<br />
(c) Given a S, then fa fS. SincefS is closed, we consider two cases. (1) fa<br />
is an isolated point and (2) fa is an accumulation point.<br />
For case (1), claim that a is an isolated point. Suppose NOT, there is a sequence<br />
<br />
x n S with x n a. Consider x n n1<br />
x : x n a x : x n a, and thus we<br />
may assume that x : x n a : a n is a infinite subset of x n <br />
n1<br />
.Sincef is<br />
monotonic, we have lim n fx n fa . SincefS is closed, we have fa fS.<br />
<strong>The</strong>refore, there exists b fS such that fa fb fa.<br />
If fb fa, then b a since f is strictly increasing. But is contradicts to that fa is<br />
isolated. On the other hand, if fb fa, then b a since f is strictly increasing. In<br />
addition, fa n fa fb implies that a n b. But is contradicts to that a n a.<br />
Hence, we have proved that a is an isolated point. So, f is sutomatically continuous at<br />
a.<br />
For case (2), suppose that fa is an accumulation point. <strong>The</strong>n Bfa; fS and<br />
Bfa; has infinite many numbers of points in fS. Choose y 1 , y 2 Bfa; fS<br />
with y 1 y 2 , then fx 1 y 1 ,andfx 2 y 2 . <strong>And</strong> thus it is similar to (a), we omit the<br />
proof.<br />
So, we have proved that f is continuous on S by (1) and (2).<br />
Remark: In (b), when we say f is monotonic on a subset of R, its image is also in R.<br />
Supplement.<br />
It should be noted that the discontinuities of a monotonic function need not be isolated.<br />
In fact, given any countable subset E of a, b, which may even be dense, we can<br />
construct a function f, monotonic on a, b, discontinuous at every point of E, and at<br />
no other point of a, b. To show this, let the points of E be arranged in a sequence x n ,<br />
n 1, 2, . . . Let c n be a sequence of positive numbers such that c n converges. Define<br />
fx c n a x b<br />
x nx<br />
Note: <strong>The</strong> summation is to be understood as follows: Sum over those indices n for whcih<br />
x n x. If there are no points x n to the left of x, the sum is empty; following the usual<br />
convention, we define it to be zero. Since absolute convergence, the order in which the<br />
terms are arranged is immaterial.<br />
<strong>The</strong>n fx is desired.<br />
<strong>The</strong> proof that we omit; the reader should see the book, Principles of Mathematical<br />
Analysis written by Walter Rudin, pp97.<br />
Metric space and fixed points<br />
4.66 Let BS denote the set of all real-valued functions which are defined and<br />
bounded on a nonempty set S. Iff BS, let<br />
f sup<br />
xS<br />
<strong>The</strong> number f is called the " sup norm " of f.<br />
|fx|.
(a) Provet that the formula df, g f g defines a metric d on BS.<br />
Proof: We prove that d isametriconBS as follows.<br />
(1) If df, g 0, i.e., f g sup xS |fx gx| 0 |fx gx| for all x S.<br />
So, we have f g on S.<br />
(2) If f g on S, then |fx gx| 0 for all x S. Thatis,f g 0 df, g.<br />
(3) Given f, g BS, then<br />
df, g f g<br />
sup|fx gx|<br />
xS<br />
sup|gx fx|<br />
xS<br />
g f<br />
dg, f.<br />
(4) Given f, g, h BS, thensince<br />
|fx gx| |fx hx| |hx gx|,<br />
we have<br />
|fx gx| <br />
sup<br />
xS<br />
|fx hx| sup|hx gx|<br />
f h h g<br />
which implies that<br />
f g sup|fx gx| f h h g.<br />
xS<br />
So,wehaveprovethatd isametriconBS.<br />
(b) Prove that the metric space BS, d is complete. Hint: If f n is a Cauchy<br />
sequence in BS, show that f n x is a Cauchy sequence of real numbers for each x in S.<br />
Proof: Let f n be a Cauchy sequence on BS, d, That is, given 0, there is a<br />
positive integer N such that as m, n N, wehave<br />
df, g f n f m sup|f n x f m x| . *<br />
xS<br />
So, for every point x S, the sequence f n x R is a Cauchy sequence. Hence, the<br />
sequence f n x is a convergent sequence, say its limit fx. It is clear that the function<br />
fx is well-defined. Let 1 in (*), then there is a positive integer N such that as<br />
m, n N, wehave<br />
|f n x f m x| 1, for all x S. **<br />
Let m , andn N, wehaveby(**)<br />
|f N x fx| 1, for all x S<br />
which implies that<br />
|fx| 1 |f N x|, for all x S.<br />
Since |f N x| BS, say its bound M, and thus we have<br />
|fx| 1 M, for all x S<br />
which implies that fx is bounded. That is, fx BS, d. Hence, we have proved that<br />
BS, d is a complete metric space.<br />
Remark: 1. We do not require that S is bounded.<br />
2. <strong>The</strong> boundedness of a function f cannot be remove since sup norm of f is finite.<br />
xS
3. <strong>The</strong> sup norm of f, often appears and is important; the reader should keep it in mind.<br />
<strong>And</strong> we will encounter it when we discuss on sequences of functions. Also, see Exercise<br />
4.67.<br />
4. Here is an important theorem, the reader can see the definition of uniform<br />
convergence in the text book, page 221.<br />
4.67 Refer to Exercise 4.66 and let CS denote the subset of BS consisting of all<br />
funtions continuous and bounded on S, where now S is a metric space.<br />
(a) Prove that CS is a closed subset of BS.<br />
Proof: Let f be an adherent point of CS, then Bf; r CS for all r 0. So,<br />
there exists a sequence f n x such that f n f as n . So, given 0, there is a<br />
positive integer N such that as n N, wehave<br />
df n , f f n f sup|f n x fx| .<br />
xS<br />
So, we have<br />
|f N x fx| . for all x S. *<br />
Given s S, and note that f N x CS, so for this , there exists a 0 such that as<br />
|x s| , x, s S, wehave<br />
|f N x f N s| . **<br />
We now prove that f is continuous at s as follows. Given 0, and let /3, then there<br />
is a 0 such that as |x s| , x, s S, wehave<br />
|fx fs| |fx f N x| |f N x f N s| |f N s fs|<br />
/3 /3 /3 by (*) and (**)<br />
.<br />
Hence, we know that f is continuous at s, and since s is arbitrary, we know that f is<br />
continuous on S.<br />
(b) Prove that the metric subspace CS is complete.<br />
Proof: By (a), we know that CS is complete since a closed subset of a complete<br />
metric space is complete.<br />
Remark: 1. In (b), we can see Exercise 4.9.<br />
2. <strong>The</strong> reader should see the text book in Charpter 9, and note that <strong>The</strong>orem 9.2 and<br />
<strong>The</strong>orem 9.3.<br />
4.68 Refer to the proof of the fixed points theorem (<strong>The</strong>orem 4.48) for notation.<br />
(a) Prove that dp, p n dx, fx n /1 .<br />
Proof: <strong>The</strong> statement is that a contraction f of a complete metric space S has a unique<br />
fixed point p. Take any point x S, and consider the sequence of iterates:<br />
x, fx, ffx,...<br />
That is, define a sequence p n inductively as follows:<br />
p 0 x, p n1 fp n n 0, 1, 2, . . .<br />
We will prove that p n converges to a fixed point of f. First we show that p n is a<br />
Cauchy sequence. Since f is a contraction (dfx, fy dx, y, 0 1 for all<br />
x, y S), we have<br />
dp n1 , p n dfp n , fp n1 dp n , p n1 ,
so, by induction, we find<br />
dp n1 , p n n dp 1 , p 0 n dx, fx.<br />
Use the triangel inequality we find, for m n,<br />
m1<br />
dp m , p n dp k1 , p k <br />
kn<br />
m1<br />
dx, fx k<br />
kn<br />
dx, fx n m<br />
1 <br />
dx, fx<br />
n<br />
1 . *<br />
Since n 0asn , we know that p n is a Cauchy sequence. <strong>And</strong> since S is<br />
complete, we have p n p S. <strong>The</strong> uniqueness is from the inequality,<br />
dfx, fy dx, y.<br />
From (*), we know that (let m )<br />
dp, p n dx, fx n<br />
1 .<br />
This inequality, which is useful in numberical work, provides an estimate for the<br />
distance from p n to the fixed point p. An example is given in (b)<br />
(b) Take fx 1 x 2/x, S 1, . Prove that f is contraction of S with<br />
2<br />
contraction constant 1/2 and fixed point p 2 . Form the sequence p n starting wth<br />
x p 0 1 and show that p n 2 2 n .<br />
Proof: First, fx fy 1 x 2/x 1 y 2/y 1 yx<br />
x y 2<br />
2 2 2 xy , then we<br />
have<br />
|fx fy| 1 y x<br />
x y 2<br />
2 xy<br />
1 2 x y 1 xy<br />
2<br />
1 2 |x y| since 1 2 xy 1.<br />
So, f is a contraction of S with contraction constant 1/2. By Fixed Point <strong>The</strong>orem, we<br />
know that there is a unique p such that fp p. Thatis,<br />
1<br />
2 p 2 p p p 2 .( 2 is not our choice since S 1, .)<br />
By (a), it is easy to know that<br />
p n 2 2 n .<br />
Remark: Here is a modefied Fixed Point <strong>The</strong>orem: Let f be function defined on a<br />
complete metric space S. If there exists a N such that df N x f N y dx, y for all<br />
x, y S, where 0 1. <strong>The</strong>n f has a unique fixed point p S.<br />
Proof: Since f N is a contraction defined on a complete metric space, with the<br />
contraction constant , with 0 1, by Fixed Point <strong>The</strong>orem, we know that there<br />
exists a unique point p S, such that
f N p p<br />
ff N p fp<br />
f N fp fp.<br />
That is, fp is also a fixed point of f N . By uniqueness, we know that fp p. In addition,<br />
if there is p S such that fp p . <strong>The</strong>n we have<br />
f 2 p fp p ,...,f N p .. p . Hence, we have p p .Thatis,f has a unique<br />
fixed point p S.<br />
4.69 Show by counterexample that the fixed-point theorem for contractions need not<br />
hold if either (a) the underlying metric space is not complete, or (b) the contraction<br />
constant 1.<br />
Solution: (a) Let f 1 1 x : 0, 1 R, then 2 |fx fy| 1 2<br />
|x y|. So,f is a<br />
contraction on 0, 1. However, it has no any fixed point since if it has, say this point p, we<br />
get 1 1 p p p 1 0, 1.<br />
2<br />
(b) Let f 1 x : 0, 1 R, then |fx fy| |x y|. So,f is a contraction with<br />
the contraction constant 1. However, it has no any fixed point since if it has, say this point<br />
p, weget1 p p 1 0, a contradiction.<br />
4.70 Let f : S S be a function from a complete metric space S, d into itself.<br />
Assume there is a real sequence a n which converges to 0 such that<br />
df n x, f n y n dx, y for all n 1 and all x, y in S, wheref n is the nth iterate of f; that<br />
is,<br />
f 1 x fx, f n1 x ff n x for n 1.<br />
Prove that f has a unique point. Hint. Apply the fixed point theorem to f m for a suitable m.<br />
Proof: Since a n 0, given 1/2, then there is a positive integer N such that as<br />
n N, wehave<br />
|a n| 1/2.<br />
Note that a n 0 for all n. Hence, we have<br />
df N x, f N y 1 dx, y for x, y in S.<br />
2<br />
That is, f N x is a contraction defined on a complete metric space, with the contraction<br />
constant 1/2. By Fixed Point <strong>The</strong>orem, we know that there exists a unique point p S,<br />
such that<br />
f N p p<br />
ff N p fp<br />
f N fp fp.<br />
That is, fp is also a fixed point of f N . By uniqueness, we know that fp p. In addition,<br />
if there is p S such that fp p . <strong>The</strong>n we have<br />
f 2 p fp p ,...,f N p .. p . Hence, we have p p .Thatis,f has a unique<br />
fixed point p S.<br />
4.71 Let f : S S be a function from a metric space S, d into itself such that<br />
dfx, fy dx, y<br />
where x y.<br />
(a) Prove that f has at most one fixed point, and give an example of such an f with no<br />
fixed point.
Proof: If p and p are fixed points of f where p p , then by hypothesis, we have<br />
dp, p dfp, fp dp, p <br />
which is absurb. So, f has at most one fixed point.<br />
Let f : 0, 1/2 0, 1/2 by fx x 2 . <strong>The</strong>n we have<br />
|fx fy| |x 2 y 2 | |x y||x y| |x y|.<br />
However, f has no fixed point since if it had, say its fixed point p, then<br />
p 2 p p 1 0, 1/2 or p 0 0, 1/2.<br />
(b) If S is compact, prove that f has exactly one fixed point. Hint. Show that<br />
gx dx, fx attains its minimum on S.<br />
Proof: Let g dx, fx, and thus show that g is continuous on a compact set S as<br />
follows. Since<br />
dx, fx dx, y dy, fy dfy, fx<br />
dx, y dy, fy dx, y<br />
2dx, y dy, fy<br />
dx, fx dy, fy 2dx, y *<br />
and change the roles of x, andy, wehave<br />
dy, fy dx, fx 2dx, y **<br />
Hence, by (*) and (**), we have<br />
|gx gy| |dx, fx dy, fy| 2dx, y for all x, y S. ***<br />
Given 0, there exists a /2 such that as dx, y , x, y S, wehave<br />
|gx gy| 2dx, y by (***).<br />
So, we have proved that g is uniformly continuous on S.<br />
So, consider min xS gx gp, p S. We show that gp 0 dp, fp. Suppose<br />
NOT, i.e., fp p. Consider<br />
df 2 p, fp dfp, p gp<br />
which contradicts to gp is the absolute maximum. Hence, gp 0 p fp. Thatis,<br />
f has a unique fixed point in S by (a).<br />
(c)Give an example with S compact in which f is not a contraction.<br />
Solution: Let S 0, 1/2, andf x 2 : S S. <strong>The</strong>n we have<br />
|x 2 y 2 | |x y||x y| |x y|.<br />
So, this f is not contraction.<br />
Remark: 1. In (b), the Choice of g is natural, since we want to get a fixed point. That<br />
is, fx x. Hence, we consider the function g dx, fx.<br />
2. Here is a exercise that makes us know more about Remark 1. Let f : 0, 1 0, 1<br />
be a continuous function, show that there is a point p such that fp p.<br />
Proof: Consider gx fx x, then g is a continuous function defined on 0, 1.<br />
Assume that there is no point p such that gp 0, that is, no such p so that fp p. So,<br />
by Intermediate Value <strong>The</strong>orem, we know that gx 0 for all x 0, 1, orgx 0<br />
for all x 0, 1. Without loss of generality, suppose that gx 0 for all x 0, 1 which<br />
is absurb since g1 f1 1 0. Hence, we know that there is a point p such that<br />
fp p.
3. Here is another proof on (b).<br />
Proof: Given any point x S, and thus consider f n x S. <strong>The</strong>n there is a<br />
convergent subsequence f nk x, say its limit p, sinceS is compact. Consider<br />
dfp, p d f limf nk x , limf nk x<br />
k k<br />
and<br />
Note that<br />
d limff nk x, limf nk x by continuity of f at p<br />
k k<br />
limdf nk1 x, f nk x 1<br />
k<br />
df nk1 x, f nk x ... df 2 f nk1 x, ff nk1 x. 2<br />
limdf 2 f nk1 x, ff nk1 x<br />
k<br />
d limf 2 f nk1 x, limff nk1 x<br />
k k<br />
d f 2 limf nk1 x , f limf nk1 x by continuity of f 2 and f at p<br />
k k<br />
df 2 p, fp. 3<br />
So, by (1)-(3), we know that<br />
fp, fp df 2 p, fp p fp<br />
by hypothesis<br />
dfx, fy dx, y<br />
where x y. Hence, f has a unique fixed point p by (a) in Exercise.<br />
Note. 1.Ifx n x, andy n y, then dx n , y n dx, y. Thatis,<br />
lim n<br />
dx n , y n d lim n<br />
x n , lim n<br />
y n .<br />
Proof: Consider<br />
dx n , y n dx n , x dx, y dy, y n and<br />
dx, y dx, x n dx n , y n dy n , y,<br />
then<br />
|dx n , y n dx, y| dx, x n dy, y n 0.<br />
So,wehaveproveit.<br />
2. <strong>The</strong> reader should compare the method with Exercise 4.72.<br />
4.72 Assume that f satisfies the condition in Exercise 4.71. If x S, letp 0 x,<br />
p n1 fp n ,andc n dp n , p n1 for n 0.<br />
(a) Prove that c n is a decreasing sequence, and let c limc n .<br />
Proof: Consider<br />
c n1 c n dp n1 , p n2 dp n , p n1 <br />
dfp n , fp n1 dp n , p n1 <br />
dp n , p n1 dp n , p n1 <br />
0,<br />
so c n is a decreasing sequence. <strong>And</strong> c n has a lower bound 0, by Completeness of R,<br />
we know that c n is a convergent sequence, say c limc n .
(b) Assume there is a subsequence p kn which converges to a point q in S. Prove that<br />
c dq, fq dfq, ffq.<br />
Deduce that q is a fixed point of f and that p n q.<br />
Proof: Since lim n p kn q, and lim n c n c, wehavelim n c kn c. So,we<br />
consider<br />
c lim n<br />
c kn<br />
lim n<br />
dp kn , p kn1 <br />
lim n<br />
dp kn , fp kn <br />
dq, fq<br />
and<br />
dp kn , p kn1 dp kn1 , p kn ... dfp kn1 , f 2 p kn1 ,<br />
we have<br />
c dq, fq lim n<br />
dfp kn1 , f 2 p kn1 dfq, f 2 q. *<br />
So, by (*) and hypoethesis<br />
dfx, fy dx, y<br />
where x y, we know that q fq c 0, in fact, this q is a unique fixed point. .<br />
In order to show that p n p, we consider (let m kn)<br />
dp m , q dp m , fq dp m1 , q .. dp kn , q<br />
So, given 0, there exists a positive integer N such that as n N, wehave<br />
dp kn , q .<br />
Hence, as m kN, wehave<br />
dp m , q .<br />
That is, p n p.
Derivatives<br />
<strong>Real</strong>-valued functions<br />
In each following exercise assume, where mecessary, a knowledge of the formulas for<br />
differentiating the elementary trigonometric, exponential, and logarithmic functions.<br />
5.1 Assume that f is said to satisfy a Lipschitz condition of order at c if there<br />
exists a positive number M (which may depend on c) and 1 ball Bc such that<br />
|fx fc| M|x c| <br />
whenever x Bc, x c.<br />
(a) Show that a function which satisfies a Lipschitz condition of order is continuous<br />
at c if 0, and has a derivative at c if 1.<br />
Proof: 1. As 0, given 0, there is a /M 1/ such that as<br />
x c , c Bc, wehave<br />
|fx fc| M|x c| M .<br />
So, we know that f is continuous at c.<br />
2. As 1, consider x Bc, andx c, wehave<br />
fx fc<br />
x c M|x c| 1 0asx c.<br />
So, we know that f has a derivative at c with f c 0.<br />
Remark: It should be note that (a) also holds if we consider the higher dimension.<br />
(b) Given an example of a function satisfying a Lipschitz condition of order 1 at c for<br />
which f c does not exist.<br />
Solution: Consider<br />
||x| |c|| |x c|,<br />
we know that |x| is a function satisfying a Lipschitz condition of order 1 at 0 for which<br />
f 0 does not exist.<br />
5.2 In each of the following cases, determine the intervals in which the function f is<br />
increasing or decreasing and find the maxima and minima (if any) in the set where each f is<br />
defined.<br />
(a) fx x 3 ax b, x R.<br />
Solution: Since f x 3x 2 a on R, we consider two cases: (i) a 0, and (ii) a 0.<br />
(i) As a 0, we know that f is increasig on R by f 0onR. In addition, if f has a<br />
local extremum at some point c, then f c 0. It implies that a 0andc 0. That is,<br />
fx x 3 b has a local extremum at 0. It is impossible since x 3 does not. So, we know<br />
that f has no maximum and minimum.<br />
(ii) As a 0, since f 3x 2 a 3 x a/3 x a/3 , we know that<br />
f x :<br />
, a/3 a/3 , a/3 a/3 , <br />
0 0 0<br />
which implies that<br />
fx :<br />
, a/3 a/3 , a/3 a/3 , <br />
<br />
. *
Hence, f is increasing on , a/3 and a/3 , , and decreasing on<br />
a/3 , a/3 . In addition, if f has a local extremum at some point c, then f c 0.<br />
It implies that c a/3 . With help of (*), we know that fx has a local maximum<br />
f a/3 and a local minimum f a/3 .<br />
(b) fx logx 2 9, |x| 3.<br />
Solution: Since f x <br />
2x , |x| 3, we know that<br />
x 2 9<br />
which implies that<br />
f x :<br />
, 3 3, <br />
0 0<br />
, 3 3, <br />
fx :<br />
.<br />
<br />
Hence, f is increasing on 3, , and decreasing on , 3. It is clear that f cannot have<br />
local extremum.<br />
(c) fx x 2/3 x 1 4 ,0 x 1.<br />
Solution: Since f x 2x13<br />
3x 1/3<br />
which implies that<br />
7x 1, 0 x 1, we know that<br />
f x :<br />
0, 1/7 1/7, 1<br />
0 0<br />
0, 1/7 1/7, 1<br />
fx :<br />
. **<br />
<br />
Hence, we know that f is increasing on 0, 1/7, and decreasing on 1/7, 1. In addition, if f<br />
has a local extremum at some interior point c, then f c 0. It implies that c 1/7. With<br />
help of (**), we know that f has a local maximum f1/7, and two local minima f0, and<br />
f1.<br />
Remark: f has the absolute maximum f1/7, and the absolute minima<br />
f0 f1 0.<br />
(d) fx sin x/x if x 0, f0 1, 0 x /2.<br />
Sulotion: Since f x cosx xtan x as 0 x /2, and f<br />
x 2<br />
0 0, in addition,<br />
f x 0asx 0 by L-Hospital Rule, we know that<br />
which implies that<br />
f x :<br />
0, /2<br />
0<br />
0, /2<br />
fx : . (***)<br />
<br />
Hence, we know that f is decreasing on 0, /2. In addition, note that there is no interior<br />
point c such that f c 0. With help of (***), we know that f has local maximum f0,<br />
and local minimum f/2.<br />
Remark: 1. Here is a proof on f 0 :Since
lim sin x 1<br />
x0 x<br />
<br />
lim<br />
x0<br />
<br />
2sin x/2 2<br />
x 0,<br />
we know that f 0 0.<br />
2. f has the absolute maximum f0, and the absolute minimum f/2.<br />
5.3 Find a polynomial f of lowest possible degree such that<br />
fx 1 a 1 , fx 2 a 2 , f x 1 b 1 , f x 2 b 2<br />
where x 1 x 2 and a 1 , a 2 , b 1 , b 2 are given real numbers.<br />
Proof: It is easy to know that the lowest degree is at most 3 since there are 4 unknows.<br />
<strong>The</strong> degree is depends on the values of a 1 , a 2 , b 1 , b 2 .<br />
5.4 Define f as follows: fx e 1/x2 if x 0, f0 0. Show that<br />
(a) f is continuous for all x.<br />
Proof: In order to show f is continuous on R, it suffices to show f is continuous at 0.<br />
Since<br />
1/x 2<br />
limfx lim<br />
x0 x0<br />
1 e 0 f0,<br />
we know that f is continuous at 0.<br />
(b) f n is continuous for all x, andf n 0 0, (n 1,2,...)<br />
Proof: In order to show f n is continuous on R, it suffices to show f n is continuous at<br />
0. Note that<br />
px<br />
lim<br />
x e x 0, where px is any real polynomial. *<br />
Claim that for x 0, we have f n x e 1/x2 P 3n 1/x, whereP 3n t is a real polynomial of<br />
degree 3n for all n 1,2,.... As n 0, f 0 x fx e 1/x2 e 1/x2 P 0 1/x, where<br />
P 0 1/x is a constant function 1. So, as n 0, it holds. Suppose that n k holds, i.e.,<br />
f k x e 1/x2 P 3k 1/x, whereP 3k t is a real polynomial of degree 3k. Consider<br />
n k 1, we have<br />
f k1 x f k x <br />
**<br />
e 1/x2 P 3k 1/x by induction hypothesis<br />
e 1/x2 2 1 x<br />
3 2<br />
P3k<br />
1 x 1 x P3k<br />
1 x .<br />
Since 2t 3 P 3k t t 2 P 3k t is a real polynomial of degree 3k 3, we define<br />
2t 3 P 3k t t 2 P 3k t P 3k3 t, and thus we have by (**)<br />
f k1 x e 1/x2 P 3k3 1/x.<br />
So, as n k 1, it holds. <strong>The</strong>refore, by Mathematical Induction, we have proved the<br />
claim.<br />
Use the claim to show that f n 0 0, (n 1,2,...) as follows. As n 0, it is trivial<br />
by hypothesis. Suppose that n k holds, i.e., f k 0 0. <strong>The</strong>n as n k 1, we have
f k x f k 0<br />
x 0<br />
fk x<br />
x by induction hypothesis<br />
P<br />
e1/x2 3k 1/x<br />
x<br />
tP 3kt<br />
(let t 1/x)<br />
<br />
e t2<br />
tP 3k t<br />
e t<br />
e t<br />
e t2 0ast ( x 0) by (*).<br />
Hence, f k1 0 0. So, by Mathematical Induction, we have proved that f n 0 0,<br />
(n 1,2,...).<br />
Since<br />
lim<br />
x0<br />
f n x lim<br />
x0<br />
e 1/x2 P 3n 1/x<br />
lim<br />
x0<br />
lim t<br />
P 3n 1/x<br />
e 1/x2<br />
P 3n t<br />
e t<br />
0by(*)<br />
f n 0,<br />
we know that f n x is continuous at 0.<br />
Remark: 1. Here is a proof on (*). Let Px be a real polynomial of degree n, and<br />
choose an even number 2N n. We consider a Taylor Expansion with Remainder as<br />
follows. Since for any x, wehave<br />
2N1<br />
e x 1k! x k e x,0<br />
2N1<br />
2N 2! x2N2 1k! x k ,<br />
k0<br />
k0<br />
then<br />
0 Px Px<br />
e x <br />
0asx <br />
2N1 1<br />
k! xk<br />
k0<br />
2N1<br />
since degPx n deg 1<br />
k0 k! xk 2N 1. By Sandwich <strong>The</strong>orem, wehave<br />
proved<br />
Px<br />
lim<br />
x e x 0.<br />
2. Here is another proof on f n 0 0, (n 1,2,...). By Exercise 5.15, it suffices to<br />
show that lim x0 f n x 0. For the part, we have proved in this exercise. So, we omit the<br />
proof. Exercise 5.15 tells us that we need not make sure that the derivative of f at 0. <strong>The</strong><br />
reader should compare with Exercise 5.15 and Exercise 5.5.<br />
3. In the future, we will encounter the exercise in Charpter 9. <strong>The</strong> Exercises tells us one<br />
important thing that the Taylor’s series about 0 generated by f converges everywhere<br />
on R, but it represents f only at the origin.<br />
5.5 Define f, g, andh as follows: f0 g0 h0 0 and, if x 0,<br />
fx sin1/x, gx x sin1/x, hx x 2 sin1/x. Show that<br />
(a) f x 1/x 2 cos1/x, ifx 0; f 0 does not exist.<br />
e t<br />
e t2
Proof: Trivially, f x 1/x 2 1<br />
cos1/x, ifx 0. Let x n <br />
2n 1 2<br />
consider<br />
fx n f0<br />
sin1/x n<br />
x n 0 x n<br />
2n 1 as n .<br />
2<br />
Hence, we know that f 0 does not exist.<br />
(b) g x sin1/x 1/x cos1/x, ifx 0; g 0 does not exist.<br />
, and thus<br />
Proof: Trivially, g 1<br />
x sin1/x 1/x cos1/x, ifx 0. Let x n ,and<br />
2n 1 2<br />
y n 1 , we know that 2n<br />
gx n g0<br />
sin 1<br />
x n 0 xn<br />
1 for all n<br />
and<br />
gy n g0<br />
sin 1<br />
y n 0 yn<br />
0 for all n.<br />
Hence, we know that g 0 does not exist.<br />
(c) h x 2x sin1/x cos1/x, ifx 0; h 0 0; lim x0 h x does not exist.<br />
Proof: Trivially, h x 2x sin1/x cos1/x, ifx 0. Consider<br />
hx h0<br />
|x sin1/x| |x| 0asx 0,<br />
x 0<br />
so we know that h 1<br />
0 0. In addition, let x n ,andy<br />
2n 1 n 1 ,wehave<br />
2n<br />
2<br />
h x n 2<br />
2n 1 and h y n 1 for all n.<br />
2<br />
Hence, we know that lim x0 h x does not exist.<br />
5.6 Derive Leibnitz’s formula for the nth derivative of the product h of two functions<br />
f and g :<br />
n<br />
h n kn f k g nk x, where kn n!<br />
k!n k! .<br />
k0<br />
Proof: We prove it by mathematical Induction. As n 1, it is clear since<br />
h f g g f. Suppose that n k holds, i.e., h k k<br />
j0<br />
jk f j g kj x. Consider<br />
n k 1, we have
h k1 h k <br />
k<br />
jk f j g kj x <br />
j0<br />
k<br />
jk f j g kj x<br />
j0<br />
k<br />
jk f j1 g kj f j g kj1 <br />
j0<br />
<br />
<br />
<br />
k1 j0<br />
jk f j1 g kj f k1 g 0<br />
k<br />
j1<br />
jk f j g kj1 f 0 g k1<br />
<br />
k1<br />
<br />
j0 jk f j1 g kj f k1 g 0<br />
k1<br />
j0<br />
k<br />
j1<br />
f j1 g kj f 0 g k1<br />
k1<br />
<br />
j0<br />
k1<br />
<br />
j0<br />
jk <br />
k1<br />
j1<br />
k<br />
j1<br />
f j1 g kj f k1 g 0 f 0 g k1<br />
f j1 g kj f k1 g 0 f 0 g k1<br />
k1<br />
jk f j1 g kj .<br />
j0<br />
So, as n k 1, it holds. Hence, by Mathematical Induction, we have proved the<br />
Leibnitz formula.<br />
Remark: We use the famous formula called Pascal <strong>The</strong>orem:n1 k1 kn n k1 ,<br />
where 0 k n.<br />
5.7 Let f and g be two functions defined and having finite third-order derivatives f x<br />
and g x for all x in R. Iffxgx 1 for all x, show that the relations in (a), (b), (c),<br />
and (d) holds at those points where the denominators are not zero:<br />
(a) f x/fx g x/gx 0.<br />
Proof: Since fxgx 1 for all x, wehavef g g f 0 for all x. By hypothesis, we<br />
have<br />
f g g f<br />
0 for those points where the denominators are not zero<br />
fg<br />
which implies that<br />
f x/fx g x/gx 0.<br />
(b) f x/f x 2f x/fx g x/g x 0.<br />
Proof: Since f g g f 0 for all x, wehavef g g f f g 2f g g f 0. By<br />
hypothesis, we have
(c) f x<br />
f x<br />
3 f xg x<br />
fxg x<br />
3 f x<br />
fx<br />
0 f g 2f g g f<br />
f g<br />
f<br />
f <br />
f<br />
f <br />
g x<br />
g x<br />
2 g<br />
g <br />
2 f<br />
f<br />
0.<br />
g<br />
g<br />
g <br />
g <br />
f<br />
f<br />
by (a).<br />
Proof: By (b), we have f g 2f g g f 0 f g 3f g 3f g fg .By<br />
hypothesis, we have<br />
0 f g 3f g 3f g fg <br />
f g<br />
(d) f x<br />
f x<br />
3 2<br />
f x<br />
f x<br />
f<br />
f <br />
3 f g <br />
f g 3 g<br />
g fg<br />
f g<br />
f<br />
f 3f g <br />
f g<br />
f<br />
f <br />
2<br />
<br />
g x<br />
g x<br />
Proof: By(c),wehave f<br />
f <br />
f <br />
f<br />
we know that f<br />
f <br />
f x<br />
f x<br />
3 2<br />
f x<br />
f x<br />
f g <br />
fg <br />
g<br />
<br />
<br />
1 2<br />
1 2<br />
3 f<br />
f<br />
3 2<br />
3 f g <br />
fg <br />
g x<br />
g x<br />
g<br />
3 f<br />
g f<br />
f <br />
f<br />
f <br />
f<br />
3 g 2<br />
2<br />
g x<br />
3 g x 2<br />
f <br />
f <br />
f <br />
f <br />
f <br />
f <br />
<br />
<br />
f 2<br />
<br />
f <br />
<br />
3g 1 g g f<br />
f g<br />
g <br />
g <br />
f 2<br />
<br />
g 2<br />
f g <br />
g x<br />
g x<br />
2<br />
.<br />
g<br />
g <br />
2<br />
.<br />
f g <br />
fg <br />
f <br />
f<br />
g <br />
g <br />
g 2<br />
g <br />
by (a).<br />
.Since<br />
g <br />
g <br />
,<br />
f <br />
f <br />
<br />
g <br />
g <br />
which implies that<br />
by (b)<br />
Note. <strong>The</strong> expression which appears on the left side of (d) is called the Schwarzian<br />
derivative of f at x.<br />
(e) Show that f and g have the same Schwarzian derivative if<br />
gx afx b/cfx d, wheread bc 0.<br />
Hint. If c 0, write af b/cf d a/c bc ad/ccf d, and apply part<br />
(d).<br />
Proof: If c 0, we have g a f b . So, we have<br />
d d
g x<br />
g x 3 2<br />
a<br />
d f x<br />
a<br />
d f x 3 2<br />
f x<br />
f x 3 2<br />
g x<br />
g x<br />
2<br />
a<br />
d f x<br />
a<br />
d f x<br />
f x<br />
f x<br />
So, f and g have the same Schwarzian derivative.<br />
If c 0, write g af b/cf d a/c bc ad/ccf d, then<br />
cg a 1 cf d 1sincead bc 0.<br />
LetG cg a, andF <br />
1<br />
bcad<br />
which implies that<br />
F <br />
F 3 2<br />
bc ad<br />
cf d, then GF 1. It implies that by (d),<br />
F <br />
F 3 2<br />
F <br />
F <br />
2<br />
.<br />
2<br />
G <br />
G 3 2<br />
F 2 c<br />
<br />
bcad f<br />
3 F c<br />
<br />
bcad f 2<br />
f<br />
f 3 2<br />
G<br />
G 3 2<br />
cg<br />
cg 3 2<br />
f 2<br />
f <br />
G 2<br />
G <br />
2<br />
G 2<br />
G <br />
<br />
<br />
cg 2<br />
cg <br />
c<br />
2<br />
bcad f<br />
c<br />
bcad f<br />
g<br />
3 g 2<br />
.<br />
g 2 g <br />
So, f and g have the same Schwarzian derivative.<br />
5.8 Let f 1 , f 2 , g 1 , g 2 be functions having derivatives in a, b. DefineF by means of<br />
the determinant<br />
Fx <br />
f 1 x f 2 x<br />
g 1 x g 2 x<br />
,ifx a, b.<br />
(a) Show that F x exists for each x in a, b and that<br />
F x <br />
f 1 x f 2 x<br />
g 1 x g 2 x<br />
<br />
f 1 x f 2 x<br />
g 1 x g 2 x<br />
.<br />
f 1 x f 2 x<br />
Proof: SinceFx <br />
f 1 g 2 f 2 g 1 ,wehave<br />
g 1 x g 2 x<br />
F f 1 g 2 f 1 g 2 f 2 g 1 f 2 g<br />
1<br />
f 1 g 2 f 2 g 1 f 1 g 2 f 2 g 1 <br />
f<br />
1 x f 2 x f 1 x f 2 x<br />
<br />
g 1 x g 2 x g 1 x g 2 x<br />
.
then<br />
(b) State and prove a more general result for nth order determinants.<br />
Proof: Claim that if<br />
F x <br />
f 11 f 12 ... f<br />
1n<br />
Fx <br />
f 21 f 22 ... f 2n<br />
<br />
... ... ... ...<br />
f n1 f n2 ... f nn<br />
f 11 f 12 ... f 1n<br />
f 21 f 22 ... f 2n<br />
,<br />
... ... ... ...<br />
f n1 f n2 ... f nn<br />
f 11 f 12 ... f 1n<br />
f 21 f 22 ... f<br />
2n<br />
...<br />
... ... ... ...<br />
f n1 f n2 ... f nn<br />
f 11 f 12 ... f 1n<br />
f 21 f 22 ... f 2n<br />
... ... ... ...<br />
f<br />
n1<br />
f n2 ... f<br />
nn<br />
We prove it by Mahematial Induction. As n 2, it has proved in (a). Suppose that n k<br />
holds, consider n k 1,<br />
f k11 f k12 ... f k1k1<br />
<br />
f 11 f 12 ... f 1k1<br />
f 21 f 22 ... f 2k1<br />
... ... ... ...<br />
<br />
f 12 ... f 1k1<br />
f 11 ... f 1k<br />
1 k11 f k11 ... ... ... ..1 k1k1 f k1k1 ... ... ...<br />
f k2 ... f kk1 f k1 ... f kk<br />
. . . . (<strong>The</strong> reader can write it down by induction hypothesis).<br />
Hence, by Mathematical Induction, wehaveprovedit.<br />
Remark: <strong>The</strong> reader should keep it in mind since it is useful in Analysis. For example,<br />
we have the following <strong>The</strong>orem.<br />
(<strong>The</strong>orem) Suppose that f, g, andh are continuous on a, b, and differentiable on<br />
a, b. <strong>The</strong>n there is a a, b such that<br />
Proof: Let<br />
f g h <br />
fa ga ha<br />
fb gb hb<br />
Fx <br />
fx gx hx<br />
fa ga ha<br />
fb gb hb<br />
then it is clear that Fx is continuous on a, b and differentiable on a, b since the<br />
operations on determinant involving addition, substraction, and multiplication without<br />
division. Consider<br />
Fa Fb 0,<br />
then by Rolle’s <strong>The</strong>orem, we know that<br />
0.<br />
,<br />
.
which implies that<br />
F 0, where a, b,<br />
f g h <br />
fa ga ha<br />
fb gb hb<br />
0. *<br />
(Application- Generalized Mean Value <strong>The</strong>orem) Suppose that f and g are<br />
continuous on a, b, and differentiable on a, b. <strong>The</strong>n there is a a, b such that<br />
fb fag f gb ga.<br />
Proof: Let hx 1, and thus by (*), we have<br />
which implies that<br />
which implies that<br />
<br />
f g <br />
fb gb<br />
f g 0<br />
fa ga 1<br />
fb gb 1<br />
<br />
0,<br />
f g <br />
fa ga<br />
fb fag f gb ga.<br />
Note: Use the similar method, we can show Mean Value <strong>The</strong>orem by letting<br />
gx x, andhx 1. <strong>And</strong> from this viewpoint, we know that Rolle’s <strong>The</strong>orem, Mean<br />
Value <strong>The</strong>orem, and Generalized Mean Value <strong>The</strong>orem are equivalent.<br />
5.9 Given n functions f 1 ,...,f n , each having nth order derivatives in a, b. A<br />
function W, called the Wronskian of f 1 ,...,f n , is defined as follows: For each x in a, b,<br />
Wx is the value of the determinant of order n whose element in the kth row and mth<br />
column is f k1 m x, wherek 1, 2, . . , n and m 1,2,...,n. [<strong>The</strong> expression f 0 m x is<br />
written for f m x.]<br />
(a) Show that W x can be obtained by replacing the last row of the determinant<br />
defining Wx by the nth derivatives f n 1 x,...,f n n x.<br />
Proof: Write<br />
Wx <br />
f 1 f 2 ... f n<br />
f 1 f 2 ... f<br />
n<br />
... ... ... ...<br />
f 1<br />
n1<br />
f 2<br />
n1<br />
... f n<br />
n1<br />
and note that if any two rows are the same, its determinant is 0; hence, by Exercise 5.8-(b),<br />
we know that<br />
,<br />
0
W x <br />
f 1 f 2 ... f n<br />
... ... ... ...<br />
f 1<br />
n2<br />
f 1<br />
n<br />
f 2<br />
n2<br />
f 2<br />
n<br />
... f n<br />
n2<br />
... f n<br />
n<br />
.<br />
(b) Assuming the existence of n constants c 1 ,...,c n , not all zero, such that<br />
c 1 f 1 x ...c n f n x 0 for every x in a, b, show that Wx 0 for each x in a, b.<br />
Proof: Since c 1 f 1 x ...c n f n x 0 for every x in a, b, wherec 1 ,...,c n , not all<br />
zero. Without loss of generality, we may assume c 1 0, we know that<br />
c 1 f k 1 x ...c n f k n x 0 for every x in a, b, where 0 k n. Hence, we have<br />
Wx <br />
f 1 f 2 ... f n<br />
f 1 f 2 ... f<br />
n<br />
... ... ... ...<br />
f 1<br />
n1<br />
f 2<br />
n1<br />
... f n<br />
n1<br />
0<br />
since the first column is a linear combination of other columns.<br />
Note. A set of functions satisfing such a relation is said to be a linearly dependent set<br />
on a, b.<br />
(c) <strong>The</strong> vanishing of the Wronskian throughout a, b is necessary, but not sufficient.<br />
for linear dependence of f 1 ,...,f n . Show that in the case of two functions, if the Wronskian<br />
vanishes throughout a, b and if one of the functions does not vanish in a, b, then they<br />
form a linearly dependent set in a, b.<br />
Proof: Let f and g be continuous and differentiable on a, b. Suppose that fx 0for<br />
all x a, b. Since the Wronskian of f and g is 0, for all x a, b, wehave<br />
fg f g 0 for all x a, b. *<br />
Since fx 0 for all x a, b, wehaveby(*),<br />
fg f g<br />
0 g <br />
0 for all x a, b.<br />
f 2 f<br />
Hence, there is a constant c such that g cf for all x a, b. Hence, f, g forms a<br />
linearly dependent set.<br />
Remark: This exercise in (b) is a impotant theorem on O.D.E. We often write (b) in<br />
other form as follows.<br />
(<strong>The</strong>orem) Letf 1 ,..,f n be continuous and differentiable on an interval I. If<br />
Wf 1 ,...,f n t 0 0forsomet 0 I, then f 1 ,..,f n is linearly independent on I<br />
Note: If f 1 ,..,f n is linearly independent on I, ItisNOT necessary that<br />
Wf 1 ,...,f n t 0 0forsomet 0 I. For example, ft t 2 |t|, andgt t 3 . It is easy to<br />
check f, g is linearly independent on 1, 1. <strong>And</strong>Wf, gt 0 for all t 1, 1.<br />
Supplement on Chain Rule and Inverse Function <strong>The</strong>orem.<br />
<strong>The</strong> following theorem is called chain rule, it is well-known that let f be defined on an<br />
open interval S, letg be defined on fS, and consider the composite function g f defined<br />
on S by the equation
g fx gfx.<br />
Assume that there is a point c in S such that fc is an interior point of fS. Iff is<br />
differentiable at c and g is differentiable at fc, then g f is differentiable at c, andwe<br />
have<br />
g f c g fcf c.<br />
We do not give a proof, in fact, the proof can be found in this text book. We will give<br />
another <strong>The</strong>orem called <strong>The</strong> Converse of Chain Rule as follows.<br />
(<strong>The</strong>ConverseofChainRule) Suppose that f, g and u are related so that<br />
fx gux. Ifux is continuous at x 0 , f x 0 exists, g ux 0 exists and not zero.<br />
<strong>The</strong>n u x 0 is defined and we have<br />
f x 0 g ux 0 u x 0 .<br />
Proof: Sincef x 0 exists, and g ux 0 exists, then<br />
fx fx 0 f x 0 x x 0 o|x x 0 | *<br />
and<br />
gux gux 0 g ux 0 ux ux 0 o|ux ux 0 |. **<br />
Since fx gux, andfx 0 gux 0 , by (*) and (**), we know that<br />
f<br />
ux ux 0 <br />
x 0 <br />
g ux 0 x x 0 o|x x 0 | o|ux ux 0 |. ***<br />
Note that since ux is continuous at x 0 , we know that o|ux ux 0 | 0asx x 0 .<br />
So, (***) means that u x 0 is defined and we have<br />
f x 0 g ux 0 u x 0 .<br />
Remark: <strong>The</strong> condition that g ux 0 is not zero is essential, for example, gx 1on<br />
1, 1 and ux |x|, wherex 0 0.<br />
(Inverse Function <strong>The</strong>roem) Suppose that f is continuous, strictly monotonic function<br />
which has an open interval I for domain and has range J. (It implies that<br />
fgx x gfx on its corresponding domain.) Assume that x 0 is a point of J such<br />
that f gx 0 is defined and is different from zero. <strong>The</strong>n g x 0 exists, and we have<br />
g x 0 1<br />
f gx 0 .<br />
Proof: Itisaresultofthe converse of chain rule note that<br />
fgx x.<br />
Mean Value <strong>The</strong>orem<br />
5.10 Given a function defined and having a finite derivative in a, b and such that<br />
lim xb<br />
fx . Prove that lim xb<br />
f x either fails to exist or is infinite.<br />
Proof: Suppose NOT, we have the existence of lim xb<br />
f x, denoted the limit by L.<br />
So, given 1, there is a 0 such that as x b , b we have<br />
|f x| |L| 1. *<br />
Consider x, a b , b with x a, then we have by (*) and Mean Value <strong>The</strong>orem,<br />
|fx fa| |f x a| where a, x<br />
|L| 1|x a|<br />
which implies that
|fx| |fa| |L| 1<br />
which contradicts to lim xb<br />
fx .<br />
Hence, lim xb<br />
f x either fails to exist or is infinite.<br />
5.11 Show that the formula in the Mean Value <strong>The</strong>orem can be written as follows:<br />
fx h fx<br />
f<br />
h<br />
x h,<br />
where 0 1.<br />
Proof: (Mean Value <strong>The</strong>orem) Letf and g be continuous on a, b and differentiable<br />
on a, b. <strong>The</strong>n there exists a a, b such that fb fa f b a. Note that<br />
a b a, where 0 1. So, we have proved the exercise.<br />
Determine as a function of x and h, and keep x 0 fixed, and find lim h0 in each<br />
case.<br />
(a) fx x 2 .<br />
Proof: Consider<br />
fx h fx<br />
h<br />
which implies that<br />
x h2 x 2<br />
h<br />
2x h 2x h f x h<br />
1/2.<br />
Hence, we know that lim h0 1/2.<br />
(b) fx x 3 .<br />
Proof: Consider<br />
fx h fx<br />
x h3 x 3<br />
3x<br />
h<br />
h<br />
2 3xh h 2 3x h 2 f x h<br />
which implies that<br />
3x 9x2 9xh 3h 2<br />
3h<br />
Since 0 1, we consider two cases. (i) x 0, (ii) x 0.<br />
(i) As x 0, since<br />
0 3x 9x2 9xh 3h 2<br />
1,<br />
3h<br />
we have<br />
<br />
3x 9x 2 9xh3h 2<br />
3h<br />
if h 0, and h is sufficiently close to 0,<br />
3x 9x 2 9xh3h 2<br />
if h 0, and h is sufficiently close to 0.<br />
3h<br />
Hence, we know that lim h0 1/2 by L-Hospital Rule.<br />
(ii) As x 0, we have<br />
<br />
3x 9x 2 9xh3h 2<br />
3h<br />
if h 0, and h is sufficiently close to 0,<br />
3x 9x 2 9xh3h 2<br />
if h 0, and h is sufficiently close to 0.<br />
3h<br />
Hence, we know that lim h0 1/2 by L-Hospital Rule.<br />
From (i) and (ii), we know that as x 0, we have lim h0 1/2.
Remark: For x 0, we can show that lim h0 3 3<br />
as follows.<br />
Proof: Since<br />
we have<br />
<br />
0 3h2<br />
3h<br />
1,<br />
3h 2<br />
3h<br />
3 h<br />
3h<br />
3 3<br />
if h 0,<br />
3h 2<br />
3 h 3 if h 0.<br />
3h 3h 3<br />
Hence, we know that lim h0 3 . 3<br />
(c) fx e x .<br />
Proof: Consider<br />
which implies that<br />
fx h fx<br />
h<br />
exh e x<br />
h<br />
e xh f x h<br />
log eh 1<br />
h<br />
.<br />
h<br />
Hence, we know that lim h0 1/2 since<br />
log eh 1<br />
h<br />
lim lim<br />
h0 h0 h<br />
lim e h h e h 1 by L-Hospital Rule.<br />
h0 he h 1<br />
Note that e h 1 h h2<br />
2 oh2 <br />
lim<br />
h0<br />
1<br />
1/2.<br />
(d) fx log x, x 0.<br />
Proof: Consider<br />
fx h fx<br />
h<br />
which implies that<br />
h o1<br />
2<br />
1 h oh<br />
2<br />
<br />
log1 h x <br />
h<br />
h<br />
x log1 h x <br />
h<br />
x log1 h x .<br />
1<br />
x h<br />
Since log1 t t t2 2<br />
ot2 ,wehave
h h<br />
x x 1<br />
lim lim<br />
h 2 x 2 o h x 2<br />
h0 h0 h h<br />
x x 1 h 2 x 2 o h x 2<br />
lim<br />
h0<br />
1<br />
lim<br />
2 h x 2 o h x 2<br />
h x 2 1 2 h x 3 o h x 3<br />
o1<br />
2<br />
1 1 h 2 x o h x <br />
h0<br />
1<br />
1/2.<br />
5.12 Take fx 3x 4 2x 3 x 2 1andgx 4x 3 3x 2 2x in <strong>The</strong>orem 5.20. Show<br />
that f x/g x is never equal to the quotient f1 f0/g1 g0 if 0 x 1. How<br />
do you reconcile this with the equation<br />
fb fa<br />
gb ga <br />
obtainable from <strong>The</strong>orem 5.20 when n 1<br />
Solution: Note that<br />
12x 2 6x 2 12 x 1<br />
4<br />
11<br />
48<br />
f x 1 <br />
g x 1 , a x 1 b,<br />
x 1<br />
4<br />
11<br />
48<br />
, where (0 1 4 11<br />
48 1).<br />
So, when we consider<br />
f1 f0<br />
g1 g0 0<br />
and<br />
f x 12x 3 6x 2 2x xg x x12x 2 6x 2,<br />
we CANNOT write f x/g x x. Otherwise, it leads us to get a contradiction.<br />
Remark: It should be careful when we use Generalized Mean Value <strong>The</strong>orem, we<br />
had better not write the above form unless we know that the denominator is not zero.<br />
5.13 In each of the following special cases of <strong>The</strong>orem 5.20, take n 1, c a, x b,<br />
and show that x 1 a b/2.<br />
(a) fx sin x, gx cosx;<br />
Proof: Since, by <strong>The</strong>orem 5.20,<br />
sin a sin b sinx 1 2cos a b sin<br />
2<br />
a 2<br />
b sinx 1 <br />
cosa cosbcosx 1 <br />
2sin a 2<br />
b sin a 2<br />
b cosx 1 ,<br />
we find that if we choose x 1 a b/2, then both are equal.<br />
(b) fx e x , gx e x .<br />
Proof: Since, by <strong>The</strong>orem 5.20,<br />
e a e b e x 1 e<br />
a<br />
e b e x 1 ,<br />
we find that if we choose x 1 a b/2, then both are equal.<br />
Can you find a general class of such pairs of functions f and g for which x 1 will always<br />
be a b/2 and such that both examples (a) and (b) are in this class
Proof: Look at the Generalized Mean Value <strong>The</strong>orem, we try to get something from<br />
the equality.<br />
fa fbg a 2<br />
b ga gbf a 2<br />
b , *<br />
if fx, andgx satisfy following two conditions,<br />
(i) f x gx and g x fx<br />
and<br />
(ii) fa fb f a 2<br />
b ga gb g a 2<br />
b ,<br />
then we have the equality (*).<br />
5.14 Given a function f defined and having a finite derivative f in the half-open<br />
interval 0 x 1 and such that |f x| 1. Define a n f1/n for n 1, 2, 3, . . . , and<br />
show that lim n a n exists.<br />
Hint. Cauchy condition.<br />
Proof: Consider n m, and by Mean Value <strong>The</strong>orem,<br />
|a n a m| |f1/n f1/m| |f p| 1 n m 1 1 n m<br />
1<br />
then a n is a Cauchy sequence since 1/n is a Cauchy sequence. Hence, we know that<br />
lim n a n exists.<br />
5.15 Assume that f has a finite derivative at each point of the open interval a, b.<br />
Assume also that lim xc f x exists and is finite for some interior point c. Prove that the<br />
value of this limit must be f c.<br />
Proof: It can be proved by Exercise 5.16; we omit it.<br />
5.16 Let f be continuous on a, b with a finite derivative f everywhere in a, b,<br />
expect possibly at c. If lim xc f x exists and has the value A, show that f c must also<br />
exist and has the value A.<br />
Proof: Consider, for x c,<br />
fx fc<br />
x c f where x, c or c, x by Mean Value <strong>The</strong>orem, *<br />
since lim xc f x exists, given 0, there is a 0 such that as x c , c c,<br />
we have<br />
A f x A .<br />
So, if we choose x c , c c in (*), we then have<br />
fx fc<br />
A x c f A .<br />
That is, f c exists and equals A.<br />
Remark: (1) Here is another proof by L-Hospital Rule. Since it is so obvious that we<br />
omit the proof.<br />
(2) We should be noted that Exercise 5.16 implies Exercise 5.15. Both methods<br />
mentioned in Exercise 5.16 are suitable for Exercise 5.15.<br />
5.17 Let f be continuous on 0, 1, f0 0, f x defined for each x in 0, 1. Prove<br />
that if f is an increasing function on 0, 1, then so is too is the function g defined by the<br />
equation gx fx/x.<br />
Proof: Sincef is an increasing function on 0, 1, we know that, for any x 0, 1
f x fx<br />
x<br />
So, let x y, wehave<br />
f x <br />
fx f0<br />
x 0<br />
f x f 0where 0, x. *<br />
gx gy g zx y, wherey z x<br />
f zz fz<br />
x y<br />
0by(*)<br />
which implies that g is an increasing function on 0, 1.<br />
5.18 Assume f has a finite derivative in a, b and is continuous on a, b with<br />
fa fb 0. Prove that for every real there is some c in a, b such that<br />
f c fc.<br />
Hint. Apply Rolle’s <strong>The</strong>orem to gxfx for a suitable g depending on .<br />
Proof: Consider gx fxe x , then by Rolle’s <strong>The</strong>orem,<br />
ga gb g ca b, wherec a, b<br />
0<br />
which implies that<br />
f c fc.<br />
z 2<br />
Remark: (1) <strong>The</strong> finding of an auxiliary function usually comes from the equation that<br />
we consider. We will give some questions around this to get more.<br />
(2)<strong>The</strong>re are some questions about finding auxiliary functions; we write it as follows.<br />
(i) Show that e e .<br />
Proof: (STUDY) Since log x is a strictly increasing on 0, , in order to show<br />
e e , it suffices to show that<br />
log e log e log e e log <br />
which implies that<br />
log e<br />
Consider fx logx<br />
x<br />
e<br />
: e, , wehave<br />
f x 1 log x<br />
x 2<br />
log <br />
.<br />
0wherex e, .<br />
So, we know that fx is strictly decreasing on e, . Hence,<br />
e e .<br />
(ii) Show that e x 1 x for all x R.<br />
loge<br />
e<br />
log<br />
<br />
.Thatis,<br />
Proof: By Taylor <strong>The</strong>orem with Remainder Term, we know that<br />
e x 1 x ec<br />
2 x2 ,forsomec.<br />
So, we finally have e x 1 x for all x R.<br />
Note: (a) <strong>The</strong> method in (ii) tells us one thing, we can give a theorem as follows. Let<br />
f C 2n1 a, b, andf 2n x exists and f 2n x 0ona, b. <strong>The</strong>n we have<br />
2n1<br />
fx f<br />
k<br />
a<br />
.<br />
k!<br />
k0
Proof: By Generalized Mean Value <strong>The</strong>orem, we complete it.<br />
(b) <strong>The</strong>re are many proofs about that e x 1 x for all x R. We list them as a<br />
reference.<br />
(b-1) Let fx e x 1 x, and thus consider the extremum.<br />
(b-2) Use Mean Value <strong>The</strong>orem.<br />
(b-3) Since e x 1 0forx 0ande x 1 0forx 0, we then have<br />
<br />
0<br />
x<br />
et 1dt 0and <br />
x<br />
0<br />
et 1dt 0.<br />
So, e x 1 x for all x R.<br />
(iii) Let f be continuous function on a, b, and differentiabel on a, b. Prove that there<br />
exists a c a, b such that<br />
f fc fa<br />
c .<br />
b c<br />
Proof: (STUDY) Since f c fcfa , we consider f<br />
bc<br />
cb c fc fa.<br />
Hence, we choose gx fx fab x, then by Rolle’s <strong>The</strong>orem,<br />
ga gb g ca b where c a, b<br />
which implies thatf c fcfa .<br />
bc<br />
(iv) Let f be a polynomial of degree n, iff 0onR, then we have<br />
f f ..f n 0onR.<br />
Proof: Let gx f f ..f n , then we have<br />
g g f 0onR since f is a polynomial of degree n. *<br />
Consider hx gxe x , then h x g x gxe x 0onR by (*). It means that h<br />
is a decreasing function on R. Since lim x hx 0bythefactg is still a polynomial,<br />
then hx 0onR. Thatis,gx 0onR.<br />
(v) Suppose that f is continuous on a, b, fa 0 fb, and<br />
x 2 f x 4xf x 2fx 0 for all x a, b. Prove that fx 0ona, b.<br />
Proof: (STUDY) Since x 2 f x 4xf x 2fx x 2 fx by Leibnitz Rule, let<br />
gx x 2 fx, then claim that gx 0ona, b.<br />
Suppose NOT, there is a point p a, b such that gp 0. Note that since fa 0,<br />
and fb 0, So, gx has an absolute maximum at c a, b. Hence, we have g c 0.<br />
By Taylor <strong>The</strong>orem with Remainder term, wehave<br />
gx gc g cx c g <br />
x c 2 ,where x, c or c, x<br />
2!<br />
gc since g c 0, and g x 0 for all x a, b<br />
0sincegc is absolute maximum.<br />
So,<br />
x 2 fx c 2 fc 0 **<br />
which is absurb since let x a in (**).<br />
(vi) Suppose that f is continuous and differentiable on 0, , and<br />
lim x f x fx 0, show that lim x fx 0.<br />
Proof: Since lim x f x fx 0, then given 0, there is M 0 such that as<br />
x M, wehave
f x fx .<br />
So, as x M, wehave<br />
e x e M e M fM e x<br />
e x fx <br />
e x e x e M e M fM .<br />
If we let e x e M e M fM gx, ande x e M e M fM hx, then we have<br />
g x e x fx h x<br />
and<br />
gM e M fM hM.<br />
Hence, for x M,<br />
e x e M e M fM gx<br />
e x fx<br />
hx e x e M e M fM<br />
It implies that, for x M,<br />
e x e M e M fM fx e x e M e M fM<br />
which implies that<br />
lim fx 0since is arbitrary.<br />
x<br />
Note: Intheprocessofproof,weusetheresultonMean Value <strong>The</strong>orem. Letf, g, and<br />
h be continuous on a, b and differentiable on a, b. Suppose fa ga ha and<br />
f x g x h x on a, b. Show that fx gx hx on a, b.<br />
Proof: By Mean Value theorem, wehave<br />
gx fx ga fa gx fx<br />
g c f c, wherec a, x.<br />
0 by hypothesis.<br />
So, fx gx on a, b. Similarly for gx hx on a, b. Hence, fx gx hx<br />
on a, b.<br />
(vii) Let fx a 1 sin x ...a n sin nx, wherea i are real for i 1, 2, . . n. Suppose that<br />
|fx| |x| for all real x. Prove that |a 1 ..na n| 1.<br />
Proof: Let x 0, and by Mean Value <strong>The</strong>orem, we have<br />
|fx f0| |fx| |a 1 sin x ...a n sin nx|<br />
|f cx|, wherec 0, x<br />
|a 1 cosc ...na n cosncx|<br />
|x| by hypothesis.<br />
So,<br />
|a 1 cosc ...na n cosnc| 1<br />
Note that as x 0 ,wehavec 0 ; hence, |a 1 ..na n| 1.<br />
Note: Here are another type:<br />
(a) |sin 2 x sin 2 y| |x y| for all x, y.<br />
(b) |tan x tan y| |x y| for all x, y , . 2 2<br />
(viii) Let f : R R be differentiable with f x c for all x, wherec 0. Show that
there is a point p such that fp 0.<br />
Proof: ByMean Vaule <strong>The</strong>orem, we have<br />
fx f0 f x 1 x f0 cx if x 0<br />
f0 f x 2 x f0 cx if x 0.<br />
So, as x large enough, we have fx 0andasx is smalle enough, we have fx 0. Since<br />
f is differentiable on R, it is continuous on R. Hence, by Intermediate Value <strong>The</strong>orem,<br />
we know that there is a point p such that fp 0.<br />
(3) Here is another type about integral, but it is worth learning. Compare with (2)-(vii).<br />
If<br />
c 0 c 1<br />
... c n<br />
2 n 1 0, where c i are real constants for i 1, 2, . . n.<br />
Prove that c 0 ...c n x n has at least one real root between 0 and 1.<br />
Proof: Suppose NOT, i.e., (i) fx : c 0 ...c n x n 0 for all x 0, 1 or (ii)<br />
fx 0 for all x 0, 1.<br />
In case (i), consider<br />
1<br />
0 fxdx c0 c 1<br />
... c n<br />
0 2 n 1 0<br />
which is absurb. Similarly for case (ii).<br />
So, we know that c 0 ...c n x n has at least one real root between 0 and 1.<br />
5.19. Assume f is continuous on a, b and has a finite second derivative f in the<br />
open interval a, b. Assume that the line segment joining the points A a, fa and<br />
B b, fb intersects the graph of f in a third point P different from A and B. Prove that<br />
f c 0forsomec in a, b.<br />
Proof: Consider a straight line equation, called gx fa fbfa x a. <strong>The</strong>n<br />
ba<br />
hx : fx gx, we knwo that there are three point x a, p and b such that<br />
ha hp hb 0.<br />
So, by Mean Value <strong>The</strong>orem twice, we know that there is a point c a, b such that<br />
h c 0<br />
which implies that f c 0sinceg is a polynomial of degree at least 1.<br />
5.20 If f has a finite third derivative f in a, b and if<br />
fa f a fb f b 0,<br />
prove that f c 0forsomec in a, b.<br />
Proof: Sincefa fb 0, we have f p 0wherep a, b by Rolle’s<br />
<strong>The</strong>orem. Since f a f p 0, we have f q 1 0whereq 1 a, p and since<br />
f p f b 0, we have f q 2 0whereq 2 p, b by Rolle’s <strong>The</strong>orem. Since<br />
f q 1 f q 2 0, we have f c 0wherec q 1 , q 2 by Rolle’s <strong>The</strong>orem.<br />
5.21 Assume f is nonnegative and has a finite third derivative f in the open interval<br />
0, 1. Iffx 0 for at least two values of x in 0, 1, prove that f c 0forsomec in<br />
0, 1.<br />
Proof: Since fx 0 for at least two values of x in 0, 1, sayfa fb 0, where<br />
a, b 0, 1. By Rolle’s <strong>The</strong>orem, we have f p 0wherep a, b. Note that f is<br />
nonnegative and differentiable on 0, 1, so both fa and fb are local minima, where a<br />
and b are interior to a, b. Hence, f a f b 0.
Since f a f p 0, we have f q 1 0whereq 1 a, p and since<br />
f p f b 0, we have f q 2 0whereq 2 p, b by Rolle’s <strong>The</strong>orem. Since<br />
f q 1 f q 2 0, we have f c 0wherec q 1 , q 2 by Rolle’s <strong>The</strong>orem.<br />
5.22 Assume f has a finite derivative in some interval a, .<br />
(a) If fx 1andf x c as x , prove that c 0.<br />
Proof: Consider fx 1 fx f y where y x, x 1 by Mean Value <strong>The</strong>orem,<br />
since<br />
lim fx 1<br />
x<br />
which implies that<br />
limfx 1 fx 0<br />
x<br />
which implies that (x y )<br />
lim<br />
x f y 0 y<br />
lim fy<br />
Since f x c as x , we know that c 0.<br />
Remark: (i) <strong>The</strong>re is a similar exercise; we write it as follows. If fx L and<br />
f x c as x , prove that c 0.<br />
Proof: By the same method metioned in (a), we complete it.<br />
(ii) <strong>The</strong> exercise tells that the function is smooth; its first derivative is smooth too.<br />
(b) If f x 1asx , prove that fx/x 1asx .<br />
Proof: Given 0, we want to find M 0 such that as x M<br />
fx<br />
x 1 .<br />
Since f x 1asx , thengiven 3 ,thereisM 0 such that as x M ,we<br />
have<br />
|f x 1| <br />
3 |f x| 1 <br />
3<br />
<br />
By Taylor <strong>The</strong>orem with Remainder Term,<br />
fx fM f x M <br />
fx x fM f 1x f M ,<br />
then for x M ,<br />
fx<br />
x 1 fM <br />
x |f 1| f M <br />
x<br />
<br />
fM <br />
x 3 1 3<br />
Choose M 0 such that as x M M ,wehave<br />
fM<br />
x <br />
3 M<br />
and x<br />
<br />
M <br />
x by (*)<br />
*<br />
**<br />
/3<br />
1 3 . ***<br />
Combine (**) with (***), we have proved that given 0, there is a M 0 such that as<br />
x M, wehave<br />
fx<br />
x 1 .<br />
That is, lim x<br />
fx<br />
x 1.
Remark: If we can make sure that fx as x , we can use L-Hopital Rule.<br />
We give another proof as follows. It suffices to show that fx as x .<br />
Proof: Since f x 1asx , thengiven 1, there is M 0 such that as<br />
x M, wehave<br />
|f x| 1 1 2.<br />
Consider<br />
fx fM f x M<br />
by Taylor <strong>The</strong>orem with Remainder Term, then<br />
lim fx since x f x is bounded for x M.<br />
(c) If f x 0asx , prove that fx/x 0asx .<br />
Proof: <strong>The</strong> method metioned in (b). We omit the proof.<br />
Remark: (i) <strong>The</strong>re is a similar exercise; we write it as follows. If f x L as x ,<br />
prove that fx/x L as x . <strong>The</strong> proof is mentioned in (b), so we omit it.<br />
(ii) It should be careful that we CANNOT use L-Hospital Rule since we may not have<br />
the fact fx as x . Hence, L-Hospital Rule cannot be used here. For example, f<br />
is a constant function.<br />
5.23 Let h be a fixed positive number. Show that there is no function f satisfying the<br />
following three conditions: f x exists for x 0, f 0 0, f x h for x 0.<br />
Proof: It is called Intermediate Value <strong>The</strong>orem for Derivatives. (Sometimes, we<br />
also call this theorem Darboux.) See the text book in <strong>The</strong>orem 5.16.<br />
(Supplement) 1. Suppose that a R, andf is a twice-differentiable real function on<br />
a, . LetM 0 , M 1 ,andM 2 are the least upper bound of |fx|, |f x|, and|f x|,<br />
respectively, on a, . Prove that M 12 4M 0 M 2 .<br />
Proof: Consider Taylor’s <strong>The</strong>orem with Remainder Term,<br />
fa 2h fa f a2h f <br />
2h 2 ,whereh 0.<br />
2!<br />
then we have<br />
f a <br />
2h 1 fa 2h fa f h<br />
which implies that<br />
Since gh : M 0<br />
h<br />
|f a| M 0<br />
h hM 2 M 1 M 0<br />
h hM 2. *<br />
hM 2 has an absolute maximum at , hence by (*), we know that<br />
M 12 4M 0 M 2 .<br />
Remark:<br />
2. Suppose that f is a twice-differentiable real function on 0, , andf is bounded on<br />
0, , andfx 0asx . Prove that f x 0asx .<br />
Proof: Since M 12 4M 0 M 2 in Supplement 1, wehaveproveit.<br />
3. Suppose that f is real, three times differentiable on 1, 1, such that f1 0,<br />
f0 0, f1 1, and f 0 0. Prove that f 3 x 3forsomex 1, 1.<br />
Proof: Consider Taylor’s <strong>The</strong>orem with Remainder Term,<br />
M 0<br />
M 2
fx f0 f 0x f 0<br />
x<br />
2!<br />
2 f3 c<br />
x<br />
3!<br />
3 ,wherec x,0 or 0, x,<br />
<strong>The</strong>n let x 1, and subtract one from another, we get<br />
f 3 c 1 f 3 c 2 6, where c 1 and c 2 in 1, 1.<br />
So,wehaveprovef 3 x 3forsomex 1, 1.<br />
5.24 If h 0andiff x exists (and is finite) for every x in a h, a h, andiff is<br />
continuous on a h, a h, show that we have:<br />
(a)<br />
fa h fa h<br />
f<br />
h<br />
a h f a h,0 1;<br />
Proof: Let gh fa h fa h, then by Mean Vaule <strong>The</strong>orem, wehave<br />
gh g0 gh<br />
g hh, where 0 1<br />
f a h f a hh<br />
which implies that<br />
fa h fa h<br />
f<br />
h<br />
a h f a h,0 1.<br />
(b)<br />
fa h 2fa fa h<br />
f<br />
h<br />
a h f a h,0 1.<br />
Proof: Let gh fa h 2fa fa h, then by Mean Vaule <strong>The</strong>orem, wehave<br />
gh g0 gh<br />
g hh, where 0 1<br />
f a h f a hh<br />
which implies that<br />
fa h 2fa fa h<br />
f<br />
h<br />
a h f a h,0 1.<br />
(c) If f a exists, show that.<br />
f fa h 2fa fa h<br />
a lim<br />
h0 h 2<br />
Proof: Since<br />
fa h 2fa fa h<br />
lim<br />
h0 h 2<br />
f<br />
lim<br />
a h f a h<br />
by L-Hospital Rule<br />
h0 2h<br />
lim<br />
h0<br />
f a h f a<br />
2h<br />
1 2 2f a since f a exists.<br />
f a.<br />
f a f a h<br />
2h<br />
Remark: <strong>The</strong>re is another proof by using Generalized Mean Value theorem.<br />
Proof: Let g 1 h fa h 2fa fa h and g 2 h h 2 , then by Generalized<br />
Mean Value theorem, we have
which implies that<br />
Hence,<br />
g 1 h g 1 0g 2 h g 1 hg 2 h g 2 0<br />
fa h 2fa fa h<br />
h 2<br />
f a h f a h<br />
2h<br />
fa h 2fa fa h<br />
lim<br />
h0 h 2<br />
f<br />
lim<br />
a h f a h<br />
h0 2h<br />
f a since f a exists.<br />
(d) Give an example where the limit of the quotient in (c) exists but where f a does<br />
not exist.<br />
Solution: (STUDY) Note that in the proof of (c) by using L-Hospital Rule. We know<br />
that |x| is not differentiable at x 0, and |x| satisfies that<br />
|0 h| |0 h|<br />
f<br />
lim<br />
0 lim<br />
0 h f 0 h<br />
h0 2h<br />
h0 2h<br />
So,letustrytofindafunctionf so that f x |x|. So, consider its integral, we know that<br />
x 2<br />
2<br />
fx <br />
if x 0<br />
x2 if x 0<br />
2<br />
Hence, we complete it.<br />
Remark: (i) <strong>The</strong>re is a related statement; we write it as follows. Suppose that f defined<br />
on a, b and has a derivative at c a, b. Ifx n a, c and y n c, b with such<br />
that x n y n 0asn . <strong>The</strong>n we have<br />
f fy<br />
c lim n fx n <br />
n y n x n<br />
.<br />
Proof: Sincef c exists, we have<br />
fy n fc f cy n c oy n c *<br />
and<br />
fx n fc f cx n c ox n c. *’<br />
If we combine (*) and (*’), we have<br />
Note that<br />
we have<br />
fy n fx n <br />
y n x n<br />
y n c<br />
y n x n<br />
lim n<br />
lim n<br />
0.<br />
which implies that, by (**)<br />
f c oy n c<br />
y n x n<br />
1and<br />
oy n c<br />
y n x n<br />
oy n c<br />
y n c<br />
x n c<br />
y n x n<br />
1 for all n,<br />
ox n c<br />
y n x n<br />
y n c<br />
y n x n<br />
ox n c<br />
x n c<br />
ox n c<br />
y n x n<br />
. **<br />
x n c<br />
y n x n
f fy<br />
c lim n fx n <br />
n y n x n<br />
.<br />
(ii) <strong>The</strong>re is a good exercise; we write it as follows. Let f C 2 a, b, andc a, b.<br />
For small |h| such that c h a, b, write<br />
fc h fc f c hhh<br />
where 0 1. Show that if f c 0, then lim h0 h 1/2.<br />
Proof: Since f C 2 a, b, byTaylor <strong>The</strong>orem with Remainder Term, we have<br />
fc h fc f ch f <br />
h<br />
2!<br />
2 ,where c, h or h, c<br />
f c hhh by hypothesis.<br />
So,<br />
f c hh f c<br />
h<br />
h f <br />
2!<br />
and let h 0, we have c by continuity of f at c. Hence,<br />
limh 1/2 since f c 0.<br />
h0<br />
Note: We can modify our statement as follows. Let f be defined on a, b, and<br />
c a, b. For small |h| such that c h a, b, write<br />
fc h fc f c hhh<br />
where 0 1. Show that if f c 0, and x x for x a h, a h, then<br />
lim h0 h 1/2.<br />
Proof: Use the exercise (c), we have<br />
f fc h 2fc fc h<br />
c lim<br />
h0 h 2<br />
f<br />
lim<br />
c hh f c hh<br />
by hypothesis<br />
h0 h<br />
f<br />
lim<br />
c hh f c hh<br />
2h since x x for x a h, a h.<br />
h0 2hh<br />
Since f c 0, we finally have lim h0 h 1/2.<br />
5.25 Let f have a finite derivatiive in a, b and assume that c a, b. Consider the<br />
following condition: For every 0, there exists a 1 ball Bc; , whose radius <br />
depends only on and not on c, such that if x Bc; , andx c, then<br />
fx fc<br />
x c f c .<br />
Show that f is continuous on a, b if this condition holds throughout a, b.<br />
Proof: Given 0, we want to find a 0 such that as dx, y , x, y a, b, we<br />
have<br />
|f x f y| .<br />
Choose any point y a, b, and thus by hypothesis, given /2, there is a 1 ball<br />
By; , whose radius depends only on and not on y, such that if x By; , and<br />
x y, then,<br />
fx fy<br />
x y f y /2 . *<br />
Note that y Bx, , so, we also have<br />
,
fx fy<br />
x y f x /2 *’<br />
Combine (*) with (*’), we have<br />
|f x f y| .<br />
Hence, we have proved f is continuous on a, b.<br />
Remark: (i) <strong>The</strong> open interval can be changed into a closed interval; it just need to<br />
consider its endpoints. That is, f is continuous on a, b if this condition holds throughout<br />
a, b. <strong>The</strong> proof is similar, so we omit it.<br />
(ii) <strong>The</strong> converse of statement in the exercise is alos true. We write it as follows. Let f <br />
be continuous on a, b, and 0. Prove that there exists a 0 such that<br />
fx fc<br />
x c f c <br />
whenever 0 |x c| , a x, c, b.<br />
Proof: Given 0, we want to find a 0 such that<br />
fx fc<br />
x c f c <br />
whenever 0 |x c| , a x, c b. Sincef is continuous on a, b, we know that f is<br />
uniformly continuous on a, b. That is, given 0, there is a 0 such that as<br />
dx, y , wehave<br />
|f x f y| . *<br />
Consider dx, c , x a, b, thenby(*),wehave<br />
fx fc<br />
x c f c |f x f c| by Mean Value <strong>The</strong>orem<br />
where dx , x . So, we complete it.<br />
Note: This could be expressed by saying that f is uniformly differentiable on a, b if f <br />
is continuous on a, b.<br />
5.26 Assume f has a finite derivative in a, b and is continuous on a, b, with<br />
a fx b for all x in a, b and |f x| 1 for all x in a, b. Prove that f has a<br />
unique fixed point in a, b.<br />
Proof: Given any x, y a, b, thus, by Mean Value <strong>The</strong>orem, wehave<br />
|fx fy| |f z||x y| |x y| by hypothesis.<br />
So, we know that f is a contraction on a complete metric space a, b. So,f has a unique<br />
fixed point in a, b.<br />
5.27 Give an example of a pair of functions f and g having a finite derivatives in 0, 1,<br />
such that<br />
but such that lim x0<br />
f x<br />
g x<br />
lim<br />
x0<br />
fx<br />
gx 0,<br />
does not exist, choosing g so that g x is never zero.<br />
Proof: Letfx sin1/x and gx 1/x. <strong>The</strong>n it is trivial for that g x is never zero.<br />
In addition, we have<br />
fx<br />
f<br />
lim 0, and lim<br />
x<br />
does not exist.<br />
x0 gx x0 g x
Remark: In this exercise, it tells us that the converse of L-Hospital Rule is NOT<br />
necessary true. Here is a good exercise very like L-Hospital Rule, but it does not! We<br />
write it as follows.<br />
Suppose that f a and g a exist with g a 0, and fa ga 0. Prove that<br />
Proof: Consider<br />
lim xa<br />
fx<br />
gx lim xa<br />
<br />
lim xa<br />
fx<br />
gx <br />
fx fa<br />
gx ga lim xa<br />
f a<br />
g a .<br />
fx fa/x a<br />
gx ga/x a<br />
f a<br />
g a since f a and g a exist with g a 0.<br />
Note: (i) It should be noticed that we CANNOT use L-Hospital Rule since the<br />
statement tells that f and g have a derivative at a, we do not make sure of the situation of<br />
other points.<br />
(ii) This holds also for complex functions. Let us recall the proof of L-Hospital Rule,<br />
we need use the order field R; however, C is not an order field. Hence, L-Hospital Rule<br />
does not hold for C. Infact,noordercanbedefinedinthecomplexfieldsincei 2 1.<br />
Supplement on L-Hospital Rule<br />
We do not give a proof about the following fact. <strong>The</strong> reader may see the book named A<br />
First Course in <strong>Real</strong> Analysis written by Protter and Morrey, Charpter 4, pp 88-91.<br />
<strong>The</strong>orem ( 0 ) Let f and g be continuous and differentiable on a, b with 0 g 0ona, b.<br />
If<br />
then<br />
lim<br />
xa <br />
lim<br />
xa<br />
fx 0 lim gx 0and<br />
*<br />
<br />
xa f x<br />
g x L,<br />
lim<br />
xa <br />
fx<br />
gx L.<br />
Remark: 1. <strong>The</strong> size of the interval a, b is of no importance; it suffices to<br />
have g 0ona, a , forsome 0.<br />
2. (*) is a sufficient condition, not a necessary condition. For example,<br />
fx x 2 ,andgx sin 1/x both defined on 0, 1.<br />
3. We have some similar results: x a ; x a; x 1/x 0 ;<br />
x 1/x 0 .<br />
<strong>The</strong>orem ( ) Let f and g be continuous and differentiable on a, b with g 0on<br />
a, b. If<br />
then<br />
lim<br />
xa <br />
lim<br />
xa<br />
fx lim gx and<br />
*<br />
<br />
xa f x<br />
g x L,
lim<br />
xa <br />
fx<br />
gx L.<br />
Remark: 1. <strong>The</strong> proof is skilled, and it needs an algebraic identity.<br />
2. We have some similar results: x a ; x a; x 1/x 0 ;<br />
x 1/x 0 .<br />
3. (*) is a sufficient condition, not a necessary condition. For example,<br />
fx x sin x, andgx x.<br />
<strong>The</strong>orem (O. Stolz) Suppose that y n , andy n is increasing. If<br />
lim<br />
x n x n1<br />
n y n y n1<br />
L, (or )<br />
then<br />
lim<br />
x n<br />
n y n<br />
L. (or )<br />
Remark: 1. <strong>The</strong> proof is skilled, and it needs an algebraic identity.<br />
2. <strong>The</strong> difference between <strong>The</strong>orem 2 and <strong>The</strong>orem 3 is that x is a continuous<br />
varibale but x n is not.<br />
<strong>The</strong>orem (Taylor <strong>The</strong>orem with Remainder) Suppose that f is a real function defined on<br />
a, b. Iff n x is continuous on a, b, and differentiable on a, b, then (let<br />
x, c a, b, with x c) there is a x, interior to the interval joining x and c such<br />
that<br />
fx P f x fn1 x<br />
n 1! x cn1 ,<br />
where<br />
n<br />
f<br />
P f x : <br />
k c<br />
x c k .<br />
k!<br />
k0<br />
Remark: 1. As n 1, it is exactly Mean Value <strong>The</strong>orem.<br />
2. <strong>The</strong> part<br />
f n1 x<br />
n 1! x cn1 : R n x<br />
is called the remainder term.<br />
3. <strong>The</strong>re are some types about remainder term. (Lagrange, Cauchy, Berstein,<br />
etc.)<br />
Lagrange<br />
R n x fn1 x<br />
n 1! x cn1 .<br />
Cauchy<br />
Berstein<br />
R n x fn1 c x c<br />
n!<br />
R n x <br />
1 n x c n1 , where 0 1.<br />
1<br />
n! c<br />
x<br />
x t n f n1 tdt
5.28 Prove the following theorem:<br />
Let f and g be two functions having finite nth derivatives in a, b. For some interior<br />
point c in a, b, assume that fc f c ... f n1 c 0, and that<br />
gc g c ... g n1 c 0, but that g n x is never zero in a, b. Show that<br />
lim xc<br />
fx<br />
gx <br />
fn c<br />
g n c .<br />
NOTE. f n and g n are not assumed to be continuous at c.<br />
Hint. Let<br />
Fx fx x cn2 f n c<br />
,<br />
n 2!<br />
define G similarly, and apply <strong>The</strong>orem 5.20 to the functions F and G.<br />
Proof: Let<br />
Fx fx <br />
fn c<br />
x cn2<br />
n 2!<br />
and<br />
Gx gx <br />
gn c<br />
x cn2<br />
n 2!<br />
then inductively,<br />
F k x f k f<br />
x <br />
n c<br />
x cn2k<br />
n 2 k!<br />
and note that<br />
F k c 0 for all k 0, 1, . . , n 3, and F n2 c f n c.<br />
Similarly for G. Hence, by <strong>The</strong>orem 5.20, wehave<br />
n2<br />
Fx <br />
k0<br />
F<br />
k<br />
k! x ck G n1 x 1 F n1 x 1 Gx <br />
k0<br />
n2<br />
G<br />
k<br />
k!<br />
x c k<br />
where x 1 between x and c, which implies that<br />
fxg n1 x 1 f n1 x 1 gx. *<br />
Note that since g n is never zero on a, b; it implies that there exists a 0 such that<br />
every g k is never zero in c , c c, wherek 0, 1, 2. . . , n. Hence, we have, by<br />
(*),<br />
lim xc<br />
fx<br />
gx lim xc<br />
lim<br />
x1 c<br />
lim<br />
x1 c<br />
<br />
f n1 x 1 <br />
g n1 x 1 <br />
f n1 x 1 f n1 c<br />
g n1 x 1 g n1 c since x c x 1 c<br />
f n1 x 1 f n1 c/x 1 c<br />
g n1 x 1 g n1 c/x 1 c<br />
fn c<br />
g n c since fn exists and g n exists 0 on a, b.<br />
Remark: (1) <strong>The</strong> hint is not correct from text book. <strong>The</strong> reader should find the<br />
difference between them.<br />
(2) Here ia another proof by L-Hospital Rule and Remark in Exercise 5.27.<br />
Proof: Since g n is never zero on a, b, it implies that there exists a 0 such that
every g k is never zero in c , c c, wherek 0, 1, 2. . . , n. So, we can apply<br />
n 1 times L-Hospital Rule methoned in Supplement, and thus get<br />
fx<br />
lim xc gx lim f n1 x<br />
xc<br />
g n1 x<br />
lim xc<br />
lim xc<br />
<br />
f n1 x f n1 c<br />
g n1 x g n1 c<br />
f n1 x f n1 c/x c<br />
g n1 x g n1 c/x c<br />
fn c<br />
g n c since fn exists and g n exists 0 on a, b.<br />
5.29 Show that the formula in Taylor’s theorem can also be written as follows:<br />
n1<br />
fx <br />
k0<br />
f<br />
k<br />
c<br />
k!<br />
x c k x cx x 1 n1<br />
n 1!<br />
f n x 1 ,<br />
where x 1 is interior to the interval joining x and c. Let1 x x 1 /x c. Show that<br />
0 1 and deduce the following form of the remainder term (due to Cauchy):<br />
1 n1 x c n<br />
f<br />
n 1!<br />
n x 1 c.<br />
Hint. Take Gt t in the proof of <strong>The</strong>orem 5.20.<br />
Proof: Let<br />
and note that<br />
n1<br />
Ft <br />
k0<br />
f<br />
k<br />
t<br />
k!<br />
x t k ,andGt t,<br />
F t <br />
fn t<br />
x tn1<br />
n 1!<br />
then by Generalized Mean Value <strong>The</strong>orem, wehave<br />
Fx FcG x 1 Gx GcF x 1 <br />
which implies that<br />
n1<br />
fx <br />
k0<br />
f<br />
k<br />
c<br />
k!<br />
So,wehaveprovethat<br />
where<br />
x c k <br />
fn x 1 <br />
n 1! x x 1 n<br />
fn x 1 c<br />
n 1!<br />
n1<br />
fx <br />
k0<br />
f<br />
k<br />
c<br />
k!<br />
x c n 1 n ,wherex 1 x 1 c.<br />
x c k R n1 x,<br />
R n1 x fn x 1 c<br />
x c n 1 n ,wherex<br />
n 1!<br />
1 x 1 c<br />
is called a Cauchy Remainder.<br />
Supplement on some questions.<br />
1. Let f be continuous on 0, 1 and differentiable on 0, 1. Suppose that f0 0and
|f x| |fx| for x 0, 1. Prove that f is constant.<br />
Proof: Given any x 1 0, 1, by Mean Value <strong>The</strong>orem and hypothesis, we know that<br />
|fx 1 f0| x 1 |f x 2 | x 1 |fx 2 |, wherex 2 0, x 1 .<br />
So, we have<br />
|fx 1 | x 1 x n|fx n1 | Mx 1 x n ,wherex n1 0, x n ,andM sup |fx|<br />
xa,b<br />
Since Mx 1 x n 0, as n , we finally have fx 1 0. Since x 1 is arbirary, we find<br />
that fx 0on0, 1.<br />
2. Suppose that g is real function defined on R, with bounded derivative, say |g | M.<br />
Fix 0, and define fx x gx. Show that f is 1-1 if is small enough. (It implies<br />
that f is strictly monotonic.)<br />
Proof: Suppose that fx fy, i.e., x gx y gy which implies that<br />
|y x| |gy gx| M|y x| by Mean Value <strong>The</strong>orem, and hypothesis.<br />
So, as is small enough, we have x y. Thatis,f is 1-1.<br />
Supplement on Convex Function.<br />
Definition(Convex Function) Let f be defined on an interval I, and given 0 1,<br />
we say that f is a convex function if for any two points x, y I,<br />
fx 1 y fx 1 fy.<br />
For example, x 2 is a convex function on R. Sometimes, the reader may see another weak<br />
definition of convex function in case 1/2. We will show that under continuity, two<br />
definitions are equivalent. In addition, it should be noted that a convex function is not<br />
necessarily continuous since we may give a jump on a continuous convex function on its<br />
boundary points, for example, fx x is a continuous convex function on 0, 1, and<br />
define a function g as follows:<br />
gx x, ifx 0, 1 and g1 g0 2.<br />
<strong>The</strong> function g is not continuous but convex. Note that if f is convex, we call f is concave,<br />
vice versa. Note that every increasing convex function of a convex function is convex. (For<br />
example, if f is convex, so is e f . ) It is clear only by definition.<br />
<strong>The</strong>orem(Equivalence) Under continuity, two definitions are equivalent.<br />
Proof: It suffices to consider if<br />
f<br />
x y<br />
2<br />
<br />
fx fy<br />
2<br />
then<br />
fx 1 y fx 1 fy for all 0 1.<br />
Since (*) holds, then by Mathematical Induction, itiseasytoshowthat<br />
f<br />
x 1 ...x 2<br />
n<br />
2 n fx 1 ...fx 2<br />
n <br />
2 n .<br />
Claim that<br />
f<br />
x 1 ...x n<br />
n fx 1 ...fx n <br />
n for all n N. **<br />
Using Reverse Induction, letx n x 1...x n1<br />
, then<br />
n1<br />
*
f<br />
x 1 ...x n<br />
n<br />
f<br />
x 1 ...x n1<br />
n x n<br />
fx n <br />
fx 1 ...fx n <br />
n by induction hypothesis.<br />
So, we have<br />
f<br />
x 1 ...x n1<br />
fx 1 ...fx n1 <br />
.<br />
n 1<br />
n 1<br />
Hence, we have proved (**). Given a rational number m/n 0, 1, where<br />
g. c. d. m, n 1; we choose x : x 1 ... x m ,andy : x m1 ... x n , thenby(**),we<br />
finally have<br />
f mx n my<br />
n n mfx n mfy<br />
n n m n fx 1 m n<br />
fy. ***<br />
Given 0, 1, then there is a sequence q n Q such that q n as n . <strong>The</strong>n<br />
by continuity and (***), we get<br />
fx 1 y fx 1 fy.<br />
Remark: <strong>The</strong> Reverse Induction is that let S N and S has two properties:(1) For<br />
every k 0, 2 k S and (2) k S and k 1 N, then k 1 S. <strong>The</strong>n S N.<br />
(Lemma) Letf be a convex function on a, b, then f is bounded.<br />
Proof: LetM maxfa, fb, then every point z I, write z a 1 b, we<br />
have<br />
fz fa 1 b fa 1 fb M.<br />
In addition, we may write z ab t, wheret is chosen so that z runs through a, b. So,<br />
2<br />
we have<br />
which implies that<br />
2f<br />
f<br />
a b<br />
2<br />
a b<br />
2<br />
1 2 f<br />
f<br />
a b<br />
2<br />
a b<br />
2<br />
t 1 2 f a b<br />
2<br />
t f a b<br />
2<br />
t<br />
t<br />
fz<br />
which implies that<br />
2f a b M : m fz.<br />
2<br />
Hence, we have proved that f is bounded above by M and bounded below by m.<br />
(<strong>The</strong>orem) Iff : I R is convex, then f satisfies a Lipschitz condition on any closed<br />
interval a, b intI. In addition, f is absolutely continuous on a, b and continuous on<br />
intI.<br />
Proof: We choose 0 so that a , b intI. By preceding lemma, we<br />
know that f is bounded, say m fx M on a , b . Given any two points x, andy,<br />
with a x y b We consider an auxiliary point z y , and a suitable <br />
then y z 1 x. So,<br />
fy fz 1 x fz 1 fx fz fx fx<br />
which implies that<br />
fy fx M m y x<br />
Change roles of x and y, we finally have<br />
<br />
M m.<br />
yx<br />
yx ,
|fy fx| K|y x|, whereK M m .<br />
That is, f satisfies a Lipschitz condition on any closed interval a, b.<br />
We call that f is absolutely continuous on a, b if given any 0, there is a 0<br />
n<br />
such that for any collection of a i , b i i1<br />
of disjoint open intervals of a, b with<br />
n<br />
b i a i , wehave<br />
k1<br />
n<br />
|fb i fa i | .<br />
k1<br />
Clearly, the choice /K meets this requirement. Finally, the continuity of f on intI is<br />
obvious.<br />
(<strong>The</strong>orem) Letf be a differentiable real function defined on a, b. Prove that f is<br />
convex if and only if f is monotonically increasing.<br />
Proof: () Suppose f is convex, and given x y, we want to show that f x f y.<br />
Choose s and t such that x u s y, then it is clear that we have<br />
fu fx<br />
u x<br />
fs fu<br />
s u <br />
Let s y , wehaveby(*)<br />
fu fx<br />
u x f y<br />
fy fs<br />
y s . *<br />
which implies that, let u x <br />
f x f y.<br />
() Suppose that f is monotonically increasing, it suffices to consider 1/2, if<br />
x y, then<br />
fx fy<br />
2<br />
f<br />
x y<br />
2<br />
<br />
fx f<br />
xy<br />
2<br />
fy f<br />
xy<br />
2 <br />
2<br />
x y<br />
4 f 1 f 2 , where 1 2<br />
0.<br />
Similarly for x y, and there is nothing to prove x y. Hence, we know that f is convex.<br />
(Corollary 1) Assume next that f x exists for every x a, b, and prove that f is<br />
convex if and only if f x 0 for all x a, b.<br />
Proof: () Suppose that f is convex, we have shown that f is monotonically increasing.<br />
So, we know that f x 0 for all x a, b.<br />
() Suppose that f x 0 for all x a, b, it implies that f is monotonically<br />
increasing. So, we know that f is convex.<br />
(Corollary 2) Let0 , then we have<br />
|y 1 | ...|y n| <br />
n<br />
1/<br />
<br />
|y 1 | ...|y n| <br />
n<br />
Proof: Letp 1, and since x p pp 1x p2 0 for all x 0, we know that<br />
fx x p is convex. So, we have (let p )<br />
by<br />
x 1 ...x n<br />
n<br />
1/<br />
/<br />
<br />
x 1 / ...x n<br />
/<br />
n *
f<br />
x 1 ...x n<br />
n fx 1 ...fx n <br />
n .<br />
Choose x i |y i | ,wherei 1, 2, . . , n. <strong>The</strong>nby(*),wehave<br />
|y 1 | ...|y n| <br />
n<br />
1/<br />
<br />
|y 1 | ...|y n| <br />
n<br />
1/<br />
.<br />
(Corollary 3) Define<br />
M r y <br />
|y 1 | r ...|y n| r<br />
n<br />
1/r<br />
,wherer 0.<br />
<strong>The</strong>n M r y is a monotonic function of r on 0, . In particular, we have<br />
M 1 y M 2 y,<br />
that is,<br />
|y 1 | ...|y n|<br />
n<br />
<br />
|y 1 | 2 ...|y n| 2<br />
n<br />
Proof: It is clear by Corollary 2.<br />
(Corollary 4) By definition of M r y in Corollary 3, wehave<br />
lim M ry |y 1 | |y<br />
r0 n| 1/n : M 0 y<br />
<br />
and<br />
lim r<br />
M r y max|y 1 |,...,|y n| : M y<br />
Proof: 1. Since M r y <br />
<strong>The</strong>orem, wehave<br />
log<br />
So,<br />
2. As r 0, we have<br />
|y 1| r ...|y n| r<br />
n 0<br />
r 0<br />
|y 1 | r ...|y n| r<br />
n<br />
max|y 1 |,...,|y n| r<br />
n<br />
1/2<br />
.<br />
1/r<br />
, taking log and thus by Mean Value<br />
1 n<br />
n |yi i1<br />
| r log|y i |<br />
1 n<br />
, where 0 r r.<br />
n |yi i1<br />
| r<br />
log<br />
lim M ry lim e<br />
r0 r0<br />
<br />
1 n<br />
lim e<br />
r0<br />
<br />
i1<br />
e n<br />
1/r<br />
n<br />
y<br />
r<br />
1 ...|yn| r<br />
n<br />
r<br />
n<br />
i1<br />
1<br />
n<br />
log y i<br />
n<br />
i1<br />
|y 1 | |y n| 1/n .<br />
y i<br />
r log y i<br />
y i<br />
r <br />
M r y max|y 1 |,...,|y n| r 1/r<br />
which implies that, by Sandwich <strong>The</strong>orem,<br />
lim r<br />
M r y max|y 1 |,...,|y n|<br />
since lim r 1 n 1/r 1.<br />
(Inequality 1) Letf be convex on a, b, and let c a, b. Define
lx fa <br />
then fx lx for all x c, b.<br />
fc fa<br />
c a<br />
x a,<br />
Proof: Consider x c, d, then c xa xc a ca<br />
xa x, wehave<br />
fc x a c fa c x a fx<br />
which implies that<br />
fc fa<br />
fx fa c a x a lx.<br />
(Inequality 2) Letf be a convex function defined on a, b. Leta s t u b,<br />
then we have<br />
ft fs<br />
t s<br />
<br />
fu fs<br />
u s<br />
<br />
fu ft<br />
u t<br />
Proof: By definition of convex, we know that<br />
fu fs<br />
fx fs u s x s, x s, u *<br />
and by inequality 1, we know that<br />
ft fs<br />
fs <br />
t s<br />
x s fx, x t, u. **<br />
So, as x t, u, by (*) and (**), we finally have<br />
ft fs fu fs<br />
t s<br />
u s .<br />
Similarly, we have<br />
Hence, we have<br />
ft fs<br />
t s<br />
fu fs<br />
u s<br />
<br />
<br />
fu fs<br />
u s<br />
fu ft<br />
u t<br />
<br />
.<br />
fu ft<br />
u t<br />
Remark: Using abvoe method, it is easy to verify that if f is a convex function on a, b,<br />
then f x and f x exist for all x a, b. In addition, if x y, wherex, y a, b, then<br />
we have<br />
f x f x f y f y.<br />
That is, f x and f x are increasing on a, b. We omit the proof.<br />
(Exercise 1) Letfx be convex on a, b, and assume that f is differentiable at<br />
c a, b, wehave<br />
lx fc f cx c fx.<br />
That is, the equation of tangent line is below fx if the equation of tangent line exists.<br />
Proof: Sincef is differentiable at c a, b, we write the equation of tangent line at c,<br />
lx fc f cx c.<br />
Define<br />
fs fc<br />
ms s c where a s c and mt <br />
then it is clear that<br />
ms f c mt<br />
which implies that<br />
.<br />
.<br />
ft fc<br />
t c<br />
where b t c,
lx fc f cx c fx.<br />
(Exercise 2) Letf : R R be convex. If f is bounded above, then f is a constant<br />
function.<br />
Proof: Suppose that f is not constant, say fa fb, wherea b. Iffb fa, we<br />
consider<br />
fx fb fb fa<br />
,wherex b<br />
x b b a<br />
which implies that as x b,<br />
fb fa<br />
fx x b fb as x <br />
b a<br />
<strong>And</strong> if fb fa, we consider<br />
fx fa fb fa<br />
x a ,wherex a<br />
b a<br />
which implies that as x a,<br />
fb fa<br />
fx x a fa as x .<br />
b a<br />
So, we obtain that f is not bouded above. So, f must be a constant function.<br />
(Exercise 3) Note that e x is convex on R. Use this to show that A. P. G. P.<br />
Proof: Sincee x e x 0onR, we know that e x is convex. So,<br />
ex 1 ...e<br />
x n<br />
n ,wherex i R, i 1, 2, . . . , n.<br />
So, let e x i y i 0, for i 1, 2, . . . , n. <strong>The</strong>n<br />
e x 1 ...xn<br />
n<br />
Vector-Valued functions<br />
y 1 y n 1/n y 1 ...y n<br />
n .<br />
5.30 If a vector valued function f is differentiable at c, prove that<br />
f c lim 1 fc h fc.<br />
h0 h<br />
Conversely, if this limit exists, prove that f is differentiable at c.<br />
Proof: Write f f 1 ,...,f n : S R R n , and let c be an interior point of S. <strong>The</strong>n if<br />
f is differentiable at c, each f k is differentiable at c. Hence,<br />
lim 1 fc h fc<br />
h0 h<br />
f<br />
lim 1 c h f 1 c<br />
,..., f nc h f n c<br />
h0 h<br />
h<br />
f<br />
lim 1 c h f 1 c f<br />
,...,lim n c h f n c<br />
h0 h<br />
h0 h<br />
f 1 c,...,f n c<br />
f c.<br />
Conversly, it is obvious by above.<br />
Remark: We give a summary about this. Let f be a vector valued function defined on<br />
S. Write f : S R n R m , c is a interior point.<br />
f f 1 ,...,f n is differentiable at c each f k is differentiable at c, *<br />
and
f f 1 ,...,f n is continuous at c each f k is continuous at c.<br />
Note: <strong>The</strong> set S can be a subset in R n , the definition of differentiation in higher<br />
dimensional space makes (*) holds. <strong>The</strong> reader can see textbook, Charpter 12.<br />
5.31 A vector-valued function f is differentiable at each point of a, b and has constant<br />
norm f. Prove that ft f t 0ona, b.<br />
Proof: Sincef, f f 2 is constant on a, b, wehavef, f 0ona, b. It implies<br />
that 2f, f 0ona, b. Thatis,ft f t 0ona, b.<br />
Remark: <strong>The</strong> proof of f, g f , g f, g is easy from definition of differentiation.<br />
So, we omit it.<br />
5.32 A vector-valued function f is never zero and has a derivative f which exists and is<br />
continuous on R. If there is a real function such that f t tft for all t, prove that<br />
there is a positive real function u and a constant vector c such that ft utc for all t.<br />
Proof: Since f t tft for all t, wehave<br />
f 1 t,...,f n t f t tft tf 1 t,...,tf n t<br />
which implies that<br />
f i t<br />
t since f is never zero. *<br />
fit<br />
Note that f i t<br />
f i<br />
is a continuous function from R to R for each i 1,2,...,n, sincef is<br />
t<br />
continuous on R, we have, by (*)<br />
f<br />
1 t<br />
ax f 1 t dt x<br />
tdt fi t f ia<br />
e t for i 1, 2, . . . , n.<br />
a<br />
e a<br />
So, we finally have<br />
ft f 1 t,...,f n t<br />
where ut e t and c <br />
e t<br />
utc<br />
f 1 a<br />
e a<br />
f 1 a fna<br />
,..., .<br />
a<br />
e e a<br />
,..., f na<br />
e a<br />
Supplement on Mean Value <strong>The</strong>orem in higher dimensional space.<br />
In the future, we will learn so called Mean Value <strong>The</strong>orem in higher dimensional<br />
space from the text book in Charpter 12. We give a similar result as supplement.<br />
Suppose that f is continuous mapping of a, b into R n and f is differentiable in a, b.<br />
<strong>The</strong>n there exists x a, b such that<br />
fb fa b af x.<br />
Proof: Letz fb fa, and define x fx z which is a real valued function<br />
defined on a, b. It is clear that x is continuous on a, b and differentiable on a, b.<br />
So, by Mean Value <strong>The</strong>orem, we know that<br />
b a xb a, wherex a, b<br />
which implies that<br />
|b a| | xb a|<br />
fb fa xb a by Cauchy-Schwarz inequality.
So, we have<br />
Partial derivatives<br />
fb fa b af x.<br />
5.33 Consider the function f defined on R 2 by the following formulas:<br />
xy<br />
fx, y if x, y 0, 0 f0, 0 0.<br />
x 2 y2 Prove that the partial derivatives D 1 fx, y and D 2 fx, y exist for every x, y in R 2 and<br />
evaluate these derivatives explicitly in terms of x and y. Also, show that f is not continuous<br />
at 0, 0.<br />
Proof: It is clear that for all x, y 0, 0, wehave<br />
D 1 fx, y y y2 x 2<br />
x 2 y 2 and D 2fx, y x x2 y 2<br />
2 x 2 y 2 . 2<br />
For x, y 0, 0, wehave<br />
fx,0 f0, 0<br />
D 1 f0, 0 lim<br />
0.<br />
x0 x 0<br />
Similarly, we have<br />
D 2 f0, 0 0.<br />
In addition, let y x and y 2x, wehave<br />
limfx, x 1/2 limfx,2x 2/5.<br />
x0 x0<br />
Hence, f is not continuous at 0, 0.<br />
Remark: <strong>The</strong> existence of all partial derivatives does not make sure the continuity of f.<br />
<strong>The</strong> trouble with partial derivatives is that they treat a function of several variables as a<br />
function of one variable at a time.<br />
5.34 Let f be defined on R 2 as follows:<br />
fx, y y x2 y 2<br />
if x,,y 0, 0, f0, 0 0.<br />
x 2 y2 Compute the first- and second-order partial derivatives of f at the origin, when they exist.<br />
Proof: For x, y 0, 0, it is clear that we have<br />
4xy<br />
D 1 fx, y <br />
3<br />
x 2 y 2 and D 2fx, y x4 4x 2 y 2 y 4<br />
2 x 2 y 2 2<br />
and for x, y 0, 0, wehave<br />
fx,0 f0, 0<br />
f0, y f0, 0<br />
D 1 f0, 0 lim<br />
0, D<br />
x0 x 0<br />
2 f0, 0 lim<br />
1.<br />
y0 y 0<br />
Hence,<br />
and<br />
D 1,2 f0, 0 lim<br />
x0<br />
D 1 fx,0 D 1 f0, 0<br />
0,<br />
x 0<br />
D 2 fx,0 D 2 f0, 0<br />
lim 2<br />
x 0<br />
x0 x does not exist,<br />
D 1 f0, y D 1 f0, 0<br />
0,<br />
y 0<br />
D 1,1 f0, 0 lim<br />
x0<br />
D 2,1 f0, 0 lim<br />
y0
D 2,2 f0, 0 lim<br />
y0<br />
D 2 f0, y D 2 f0, 0<br />
y 0<br />
lim<br />
y0<br />
0 y 0.<br />
Remark: We do not give a detail computation, but here are answers. Leave to the<br />
reader as a practice. For x, y 0, 0, wehave<br />
D 1,1 fx, y 4y3 y 2 3x 2 <br />
x 2 y 2 3 ,<br />
and<br />
complex-valued functions<br />
D 1,2 fx, y 4xy2 3x 2 y 2 <br />
x 2 y 2 3 ,<br />
D 2,1 fx, y 4xy2 3x 2 y 2 <br />
x 2 y 2 3 ,<br />
D 2,2 fx, y 4x2 yy 2 3x 2 <br />
x 2 y 2 3 .<br />
5.35 Let S be an open set in C and let S be the set of complex conjugates z, where<br />
z S. Iff is defined on S, defineg on S as follows: gz fz, the complex conjugate<br />
of fz. Iff is differentiable at c, prove that g is differentiable at c and that g c f c.<br />
Proof: Sincec S, we know that c is an interior point. Thus, it is clear that c is also an<br />
interior point of S . Note that we have<br />
the conjugate of<br />
fz fc<br />
z c<br />
z<br />
f fc<br />
z c<br />
gz gc<br />
<br />
by gz fz.<br />
z c<br />
Note that z c z c, so we know that if f is differentiable at c, prove that g is<br />
differentiable at c and that g c f c.<br />
5.36 (i) In each of the following examples write f u iv and find explicit formulas<br />
for ux, y and vx, y : ( <strong>The</strong>se functions are to be defined as indicated in Charpter 1.)<br />
(a) fz sin z,<br />
Solution: Since e iz cosz i sin z, we know that<br />
sin z 1 2 ey e y sin x ie y e y cosx<br />
from sin z eiz e iz<br />
2i<br />
and<br />
. So, we have<br />
ux, y ey e y sin x<br />
2<br />
vx, y ey e y cosx<br />
.<br />
2<br />
(b) fz cosz,<br />
Solution: Sincee iz cosz i sin z, we know that<br />
cosz 1 2 ey e y cosx e y e y sin x<br />
from cos z eiz e iz<br />
2<br />
. So, we have
ux, y ey e y cosx<br />
2<br />
and<br />
vx, y ey e y sin x<br />
.<br />
2<br />
(c) fz |z|,<br />
Solution: Since|z| x 2 y 2 1/2 , we know that<br />
ux, y x 2 y 2 1/2<br />
and<br />
vx, y 0.<br />
(d) fz z,<br />
Solution: Since z x iy, we know that<br />
ux, y x<br />
and<br />
vx, y y.<br />
(e) fz arg z, (z 0),<br />
Solution: Since argz R, we know that<br />
ux, y argx iy<br />
and<br />
vx, y 0.<br />
(f) fz Log z, (z 0),<br />
Solution: Since Log z log|z| i argz, we know that<br />
ux, y logx 2 y 2 1/2<br />
and<br />
vx, y argx iy.<br />
and<br />
(g) fz e z2 ,<br />
Solution: Since e z2<br />
e x2 y 2 i2xy<br />
, we know that<br />
ux, y e x2 y 2 cos2xy<br />
vx, y e x2 y 2 sin2xy.<br />
(h) fz z ,( complex, z 0).<br />
Solution: Since z e Log z , thenwehave(let 1 i 2 )<br />
z e 1i 2 log|z|i arg z<br />
e 1 log|z| 2 arg zi 2 log|z| 1 arg z<br />
.<br />
So, we know that<br />
ux, y e 1 log|z| 2 arg z<br />
cos 2 log|z| 1 arg z<br />
e 1 logx 2 y 2 1/2 2 argxiy<br />
cos 2 logx 2 y 2 1/2 1 argx iy
and<br />
vx, y e 1 log|z| 2 arg z<br />
sin 2 log|z| 1 arg z<br />
e 1 logx 2 y 2 1/2 2 argxiy<br />
sin 2 logx 2 y 2 1/2 1 argx iy .<br />
(ii) Show that u and v satisfy the Cauchy -Riemanns equation for the following values<br />
of z :Allz in (a), (b), (g); no z in (c), (d), (e); all z except real z 0in(f),(h).<br />
So,<br />
and<br />
Proof: (a)sinz u iv, where<br />
ux, y ey e y sin x<br />
2<br />
u x v y ey e y cosx<br />
2<br />
and vx, y ey e y cosx<br />
2<br />
for all z x iy<br />
u y v x ey e y sin x<br />
for all z x iy.<br />
2<br />
(b) cos z u iv, where<br />
ux, y ey e y cosx<br />
and vx, y ey e y sin x<br />
2<br />
2<br />
So,<br />
u x v y ey e y sin x<br />
for all z x iy.<br />
2<br />
and<br />
u y v x ey e y cosx<br />
for all z x iy.<br />
2<br />
(c) |z| u iv, where<br />
ux, y x 2 y 2 1/2 and vx, y 0.<br />
So,<br />
u x xx 2 y 2 1/2 v y 0ifx 0, y 0.<br />
and<br />
u y yx 2 y 2 1/2 v x 0ifx 0, y 0.<br />
So, we know that no z makes Cauchy-Riemann equations hold.<br />
(d) z u iv, where<br />
ux, y x and vx, y y.<br />
So,<br />
u x 1 1 v y .<br />
So, we know that no z makes Cauchy-Riemann equations hold.<br />
(e) arg z u iv, where<br />
ux, y argx 2 y 2 1/2 and vx, y 0.<br />
Note that<br />
.<br />
.
ux, y <br />
(1) arctany/x, ifx 0, y R<br />
(2) /2, if x 0, y 0<br />
(3) arctany/x , ifx 0, y 0<br />
(4) arctany/x , ifx 0, y 0<br />
(5) /2, if x 0, y 0.<br />
and<br />
v x v y 0.<br />
So, we know that by (1)-(5), for x, y 0, 0<br />
y<br />
u x <br />
x 2 y 2<br />
and for x, y x, y : x 0, y 0, wehave<br />
u y x<br />
x 2 y . 2<br />
Hence, we know that no z makes Cauchy-Riemann equations hold.<br />
Remark: We can give the conclusion as follows:<br />
y<br />
arg z x<br />
for x, y 0, 0<br />
x 2 y 2<br />
and<br />
arg z y<br />
x for x, y x, y : x 0, y 0.<br />
x 2 y2 (f) Log z u iv, where<br />
ux, y logx 2 y 2 1/2 and vx, y argx 2 y 2 1/2 .<br />
Since<br />
u x x<br />
x 2 y and u y<br />
2 y <br />
x 2 y 2<br />
and<br />
y<br />
v x <br />
x 2 y for x, y 0, 0 and v 2 y x for x, y x, y : x 0, y 0,<br />
x 2 y2 we know that all z except real z 0 make Cauchy-Riemann equations hold.<br />
Remark: Log z is differentiable on C x, y : x 0, y 0 since Cauchy-Riemann<br />
equations along with continuity of u x iv x ,andu y iv y .<br />
(g) e z2 u iv, where<br />
ux, y e x2 y 2 cos2xy and vx, y e x2 y 2 sin2xy.<br />
So,<br />
u x v y 2e x2 y 2 xcos2xy ysin 2xy for all z x iy.<br />
and<br />
u y v x 2e x2 y 2 ycos2xy xsin 2xy for all z x iy.<br />
Hence, we know that all z make Cauchy-Riemann equations hold.<br />
(h) Since z e Log z ,ande z is differentiable on C, we know that, by the remark of (f),<br />
we know that z is differentiable for all z except real z 0. So, we know that all z except<br />
real z 0 make Cauchy-Riemann equations hold.<br />
( In part (h), the Cauchy-Riemann equations hold for all z if is a nonnegative integer,
and they hold for all z 0if is a negative integer.)<br />
Solution: It is clear from definition of differentiability.<br />
(iii) Compute the derivative f z in (a), (b), (f), (g), (h), assuming it exists.<br />
Solution: Sincef z u x iv x , if it exists. So, we know all results by (ii).<br />
5.37 Write f u iv and assume that f has a derivative at each point of an open disk D<br />
centered at 0, 0. Ifau 2 bv 2 is constant on D for some real a and b, not both 0. Prove<br />
that f is constant on D.<br />
Proof: Letau 2 bv 2 be constant on D. We consider three cases as follows.<br />
1. As a 0, b 0, then we have<br />
v 2 is constant on D<br />
which implies that<br />
vv x 0.<br />
If v 0onD, it is clear that f is constant.<br />
If v 0onD, that is v x 0onD. So, we still have f is contant.<br />
2. As a 0, b 0, then it is similar. We omit it.<br />
3. As a 0, b 0, Taking partial derivatives we find<br />
auu x bvv x 0onD. 1<br />
and<br />
auu y bvv y 0onD.<br />
By Cauchy-Riemann equations the second equation can be written as we have<br />
auv x bvu x 0onD. 2<br />
We consider 1v x 2u x and 1u x 2v x , then we have<br />
bvv x2 u x2 0 3<br />
and<br />
auv x2 u x2 0 4<br />
which imply that<br />
au 2 bv 2 v x2 u x2 0. 5<br />
If au 2 bv 2 c, constant on D, wherec 0, then v x2 u x2 0. So, f is constant.<br />
If au 2 bv 2 c, constant on D, wherec 0, then if there exists x, y such that<br />
v x2 u x2 0, then by (3) and (4), ux, y vx, y 0. By continuity of v x2 u x2 , we know<br />
that there exists an open region S D such that u v 0onS. Hence, by Uniqueness<br />
<strong>The</strong>orem, we know that f is constant.<br />
Remark: In complex theory, the Uniqueness theorem is fundamental and important.<br />
<strong>The</strong> reader can see this from the book named <strong>Complex</strong> Analysis by Joseph Bak and<br />
Donald J. Newman.
Functions of Bounded Variation and Rectifiable Curves<br />
Functions of bounded variation<br />
6.1 Determine which of the follwoing functions are of bounded variation on 0, 1.<br />
(a) fx x 2 sin1/x if x 0, f0 0.<br />
(b) fx x sin1/x if x 0, f0 0.<br />
Proof: (a) Since<br />
f x 2x sin1/x cos1/x for x 0, 1 and f 0 0,<br />
we know that f x is bounded on 0, 1, in fact, |f x| 3on0, 1. Hence, f is of<br />
bounded variation on 0, 1.<br />
1<br />
(b) First, we choose n 1beanevenintegersothat 1, and thus consider a<br />
2 n1<br />
partition P 0 x 0 , x 1 1 <br />
we have<br />
2<br />
n2<br />
<br />
k1<br />
, x 2 1<br />
2 2<br />
,..., x n 1<br />
n 2<br />
|f k | 2 2 <br />
n<br />
k1<br />
, x n1 <br />
1<br />
n1 2<br />
1/k .<br />
, x n2 1 , then<br />
Since 1/k diverges to , we know that f is not of bounded variation on 0, 1.<br />
6.2 A function f, defined on a, b, is said to satisfy a uniform Lipschitz condition of<br />
order 0ona, b if there exists a constant M 0 such that |fx fy| M|x y| for<br />
all x and y in a, b. (Compare with Exercise 5.1.)<br />
(a) If f is such a function, show that 1 implies f is constant on a, b, whereas<br />
1 implies f is of bounded variation a, b.<br />
Proof: As 1, we consider, for x y, wherex, y a, b,<br />
|fx fy|<br />
0 M|x y| 1 .<br />
|x y|<br />
Hence, f x exists on a, b, and we have f x 0ona, b. So, we know that f is<br />
constant.<br />
As 1, consider any partition P a x 0 , x 1 ,..., x n b, wehave<br />
n<br />
<br />
k1<br />
n<br />
|f k | M |x k1 x k | Mb a.<br />
k1<br />
That is, f is of bounded variation on a, b.<br />
(b) Give an example of a function f satisfying a uniform Lipschitz condition of order<br />
1ona, b such that f is not of bounded variation on a, b.<br />
Proof: First, note that x satisfies uniform Lipschitz condition of order , where<br />
1<br />
0 1. Choosing 1 such that 1 and let M k1<br />
since the series<br />
k <br />
converges. So, we have 1 <br />
1 1<br />
.<br />
M k1 k <br />
Define a function f as follows. We partition 0, 1 into infinitely many subsintervals.<br />
Consider<br />
x 0 0, x 1 x 0 <br />
M 1 1 1 , x 2 x 1 <br />
M 1 1 2 ,..., x n x n1 <br />
M 1 1 n ,.... <br />
<strong>And</strong> in every subinterval x i , x i1 ,wherei 0,1,...., we define
fx x x i x i1<br />
2<br />
then f is a continuous function and is not bounded variation on 0, 1 since k1<br />
<br />
,<br />
1<br />
2M<br />
diverges.<br />
In order to show that f satisfies uniform Lipschitz condition of order , we consider<br />
three cases.<br />
(1) If x, y x i , x i1 ,andx, y x i , x ix i1<br />
2<br />
or x, y x ix i1<br />
2<br />
, x i1 , then<br />
|fx fy| |x y | |x y| .<br />
(2) If x, y x i , x i1 ,andx x i , x ix i1<br />
or y x ix i1<br />
, x<br />
2 2 i1 , then there is a<br />
z x i , x ix i1<br />
such that fy fz. So,<br />
2<br />
|fx fy| |fx fz| |x z | |x z| |x y| .<br />
(3) If x x i , x i1 and y x j , x j1 ,wherei j.<br />
If x x i , x ix i1<br />
, then there is a z x<br />
2 i , x ix i1<br />
such that fy fz. So,<br />
2<br />
|fx fy| |fx fz| |x z | |x z| |x y| .<br />
Similarly for x x ix i1<br />
, x<br />
2 i1 .<br />
Remark: Here is another example. Since it will use Fourier <strong>The</strong>ory, we do not give a<br />
proof. We just write it down as a reference.<br />
<br />
ft <br />
k1<br />
cos3k t<br />
3 k .<br />
(c) Give an example of a function f which is of bounded variation on a, b but which<br />
satisfies no uniform Lipschitz condition on a, b.<br />
Proof: Since a function satisfies uniform Lipschitz condition of order 0, it must be<br />
continuous. So, we consider<br />
x if x a, b<br />
fx <br />
b 1ifx b.<br />
Trivially, f is not continuous but increasing. So, the function is desired.<br />
Remark: Here is a good problem, we write it as follows. If f satisfies<br />
|fx fy| K|x y| 1/2 for x 0, 1, wheref0 0.<br />
define<br />
fx<br />
x 1/3<br />
if x 0, 1<br />
gx <br />
0ifx 0.<br />
<strong>The</strong>n g satisfies uniform Lipschitz condition of order 1/6.<br />
Proof: Note that if one of x, andy is zero, the result is trivial. So, we may consider<br />
0 y x 1 as follows. Consider<br />
|gx gy| <br />
fx<br />
x 1/3<br />
<br />
<br />
fx<br />
x 1/3<br />
fx<br />
x 1/3<br />
fy<br />
y 1/3<br />
fy<br />
x 1/3<br />
fy<br />
x 1/3<br />
fy fy<br />
x 1/3 x 1/3<br />
fy<br />
y 1/3<br />
1<br />
k <br />
fy<br />
y 1/3 . *
For the part<br />
fx<br />
x fy 1 |fx fy|<br />
1/3 x 1/3 x1/3 K<br />
x |x 1/3 y|1/2 by hypothesis<br />
K|x y| 1/2 |x y| 1/3 since x x y 0<br />
K|x y| 1/6 . A<br />
fy<br />
For another part fy , we consider two cases.<br />
x 1/3 y 1/3<br />
(1) x 2y which implies that x x y y 0,<br />
fy<br />
x 1/3<br />
fy<br />
y 1/3<br />
|fy|<br />
|fy|<br />
x 1/3 y 1/3<br />
xy 1/3<br />
x y 1/3<br />
xy 1/3 since |x 1/3 y 1/3 | |x y| 1/3 for all x, y 0<br />
|fy| x 1/3<br />
xy 1/3 since x y 1/3 x 1/3<br />
|fy| 1<br />
y 1/3<br />
K |y|1/2<br />
|y| 1/3<br />
K|y| 1/6<br />
by hypothesis<br />
K|x y| 1/6 since y x y. B<br />
(2) x 2y which implies that x y x y 0,<br />
fy<br />
x 1/3<br />
fy<br />
y 1/3<br />
|fy|<br />
|fy|<br />
|fy|<br />
x 1/3 y 1/3<br />
xy 1/3<br />
x y 1/3<br />
xy 1/3 since |x 1/3 y 1/3 | |x y| 1/3 for all x, y 0<br />
x y 1/3<br />
y 2/3<br />
K|y| 1/2 x y 1/3<br />
y 2/3<br />
K|y| 1/6 |x y| 1/3<br />
since x y<br />
by hypothesis<br />
K|x y| 1/6 |x y| 1/3 since y x y<br />
K|x y| 1/6 . C<br />
So, by (A)-(C), (*) tells that g satisfies uniform Lipschitz condition of order 1/6.<br />
Note: Hereisageneralresult.Let0 2. Iff satisfies<br />
|fx fy| K|x y| for x 0, 1, wheref0 0.<br />
define
fx<br />
x <br />
if x 0, 1<br />
gx <br />
0ifx 0.<br />
<strong>The</strong>n g satisfies uniform Lipschitz condition of order . <strong>The</strong> proof is similar, so we<br />
omit it.<br />
6.3 Show that a polynomial f is of bounded variation on every compact interval a, b.<br />
Describe a method for finding the total variation of f on a, b if the zeros of the derivative<br />
f are known.<br />
Proof: If f is a constant, then the total variation of f on a, b is zero. So, we may<br />
assume that f is a polynomial of degree n 1, and consider f x 0 by two cases as<br />
follows.<br />
(1) If there is no point such that f x 0, then by Intermediate Value <strong>The</strong>orem of<br />
Differentiability, we know that f x 0ona, b, orf x 0ona, b. So, it implies<br />
that f is monotonic. Hence, the total variation of f on a, b is |fb fa|.<br />
(2) If there are m points such that f x 0, say<br />
a x 0 x 1 x 2 ... x m b x m1 , where 1 m n, then we know the monotone<br />
property of function f. So, the total variation of f on a, b is<br />
m1<br />
|fx i fx i1 |.<br />
i1<br />
Remark: Here is another proof. Let f be a polynomial on a, b, then we know that f is<br />
bounded on a, b since f is also polynomial which implies that it is continuous. Hence, we<br />
know that f is of bounded variation on a, b.<br />
6.4 A nonempty set S of real-valued functions defined on an interval a, b is called a<br />
linear space of functions if it has the following two properties:<br />
(a) If f S, then cf S for every real number c.<br />
(b) If f S and g S, then f g S.<br />
<strong>The</strong>orem 6.9 shows that the set V of all functions of bounded variation on a, b is a<br />
linear space. If S is any linear space which contains all monotonic functions on a, b,<br />
prove that V S. This can be described by saying that the functions of bounded<br />
variation form the samllest linear space containing all monotonic functions.<br />
Proof: It is directlt from <strong>The</strong>orem 6.9 and some facts in Linear Algebra. We omit the<br />
detail.<br />
6.5 Let f be a real-valued function defined on 0, 1 such that f0 0, fx x for all<br />
x, andfx fy whenever x y. LetA x : fx x. Prove that sup A A, andthat<br />
f1 1.<br />
Proof: Note that since f0 0, A is not empty. Suppose that sup A : a A, i.e.,<br />
fa a since fx x for all x. So, given any n 0, then there is a b n A such that<br />
a n b n . *<br />
In addition,<br />
b n fb n since b n A. **<br />
So,by(*)and(**),wehave(let n 0 ),<br />
a fa fa since f is monotonic increasing.<br />
which contradicts to fa a. Hence, we know that sup A A.
Claim that 1 sup A. Suppose NOT, thatis,a 1. <strong>The</strong>n we have<br />
a fa f1 1.<br />
Since a sup A, consider x a, fa, then<br />
fx x<br />
which implies that<br />
fa a<br />
which contradicts to a fa. So, we know that sup A 1. Hence, we have proved that<br />
f1 1.<br />
Remark: <strong>The</strong> reader should keep the method in mind if we ask how to show that<br />
f1 1 directly. <strong>The</strong> set A is helpful to do this. Or equivalently, let f be strictly increasing<br />
on 0, 1 with f0 0. If f1 1, then there exists a point x 0, 1 such that fx x.<br />
6.6 If f is defined everywhere in R 1 , then f is said to be of bounded variation on<br />
, if f is of bounded variation on every finite interval and if there exists a positive<br />
number M such that V f a, b M for all compact interval a, b. <strong>The</strong> total variation of f on<br />
, is then defined to be the sup of all numbers V f a, b, a b , and<br />
denoted by V f , . Similar definitions apply to half open infinite intervals a, and<br />
, b.<br />
(a) State and prove theorems for the inifnite interval , analogous to the<br />
<strong>The</strong>orems 6.7, 6.9, 6.10, 6.11, and 6.12.<br />
(<strong>The</strong>orem 6.7*) Letf : R R be of bounded variaton, then f is bounded on R.<br />
Proof: Given any x R, then x 0, a or x a,0. Ifx 0, a, then f is bounded<br />
on 0, a with<br />
|fx| |f0| V f 0, a |f0| V f , .<br />
Similarly for x a,0.<br />
(<strong>The</strong>orem 6.9*) Assume that f, andg be of bounded variaton on R, thensoarethier<br />
sum, difference, and product. Also, we have<br />
V fg , V f , V g , <br />
and<br />
V fg , AV f , BV g , ,<br />
where A sup xR |gx| and B sup xR |fx|.<br />
Proof: For sum and difference, given any compact interval a, b, wehave<br />
V fg a, b V f a, b V g a, b,<br />
V f , V g , <br />
which implies that<br />
V fg , V f , V g , .<br />
For product, given any compact interval a, b, wehave(letAa, b sup xa,b |gx|,<br />
and Ba, b sup xa,b |fx|),<br />
V fg a, b Aa, bV f a, b Ba, bV g a, b<br />
AV f , BV g , <br />
which implies that
V fg , AV f , BV g , .<br />
(<strong>The</strong>orem 6.10*) Letf be of bounded variation on R, and assume that f is bounded<br />
away from zero; that is, suppose that there exists a positive number m such that<br />
0 m |fx| for all x R. <strong>The</strong>n g 1/f is also of bounded variation on R, and<br />
V g , V f, <br />
m 2 .<br />
Proof: Given any compacgt interval a, b, wehave<br />
which implies that<br />
V g a, b V fa, b<br />
m 2<br />
V f, <br />
m 2<br />
V g , V f, <br />
.<br />
m 2<br />
(<strong>The</strong>orem 6.11*) Letf be of bounded variation on R, and assume that c R. <strong>The</strong>n f is<br />
of bounded variation on , c and on c, and we have<br />
V f , V f , c V f c, .<br />
Proof: Given any a compact interval a, b such that c a, b. <strong>The</strong>n we have<br />
V f a, b V f a, c V f c, b.<br />
Since<br />
V f a, b V f , <br />
which implies that<br />
V f a, c V f , and V f c, b V f , ,<br />
we know that the existence of V f , c and V f c, . Thatis,f is of bounded variation on<br />
, c and on c, .<br />
Since<br />
V f a, c V f c, b V f a, b V f , <br />
which implies that<br />
V f , c V f c, V f , , *<br />
and<br />
V f a, b V f a, c V f c, b V f , c V f c, <br />
which implies that<br />
V f , V f , c V f c, , **<br />
we know that<br />
V f , V f , c V f c, .<br />
(<strong>The</strong>orem 6.12*) Letf be of bounded variation on R. LetVx be defined on , x as<br />
follows:<br />
Vx V f , x if x R, andV 0.<br />
<strong>The</strong>n (i) V is an increasing function on , and (ii) V f is an increasing function on<br />
, .<br />
Proof: (i)Letx y, then we have Vy Vx V f x, y 0. So, we know that V is<br />
an increasing function on , .<br />
(ii) Let x y, then we have V fy V fx V f x, y fy fx 0. So,
we know that V f is an increasing function on , .<br />
(b) Show that <strong>The</strong>orem 6.5 is true for , if ”monotonic” is replaced by ”bounded<br />
and monotonic.” State and prove a similar modefication of <strong>The</strong>orem 6.13.<br />
(<strong>The</strong>orem 6.5*) Iff is bounded and monotonic on , , then f is of bounded<br />
variation on , .<br />
Proof: Given any compact interval a, b, then we have V f a, b exists, and we have<br />
V f a, b |fb fa|, sincef is monotonic. In addition, since f is bounded on R, say<br />
|fx| M for all x, we know that 2M is a upper bounded of V f a, b for all a, b. Hence,<br />
V f , exists. That is, f is of bounded variation on R.<br />
(<strong>The</strong>orem 6.13*) Letf be defined on , , then f is of bounded variation on<br />
, if, and only if, f can be expressed as the difference of two increasing and<br />
bounded functions.<br />
Proof: Suppose that f is of bounded variation on , , then by <strong>The</strong>orem 6.12*, we<br />
know that<br />
f V V f,<br />
where V and V f are increasing on , . In addition, since f is of bounded variation<br />
on R, we know that V and f is bounded on R which implies that V f is bounded on R. So,<br />
we have proved that if f is of bounded variation on , then f can be expressed as the<br />
difference of two increasing and bounded functions.<br />
Suppose that f can be expressed as the difference of two increasing and bounded<br />
functions, say f f 1 f 2 , <strong>The</strong>n by <strong>The</strong>orem 6.9*, and<strong>The</strong>orem 6.5*, we know that f is of<br />
bounded variaton on R.<br />
Remark: <strong>The</strong> representation of a function of bounded variation as a difference of two<br />
increasing and bounded functions is by no mean unique. It is clear that <strong>The</strong>orem 6.13*<br />
also holds if ”increasing” is replaced by ”strictly increasing.” For example,<br />
f f 1 g f 2 g, whereg is any strictly increasing and bounded function on R. One<br />
of such g is arctan x.<br />
6.7 Assume that f is of bounded variation on a, b and let<br />
P x 0 , x 1 ,...,x n þa, b.<br />
As usual, write f k fx k fx k1 , k 1,2,...,n. Define<br />
AP k : f k 0, BP k : f k 0.<br />
<strong>The</strong> numbers<br />
and<br />
p f a, b sup<br />
f k : P þa, b<br />
kAP<br />
n f a, b sup<br />
|f k | : P þa, b<br />
kBP<br />
are called respectively, the positive and negative variations of f on a, b. For each x in<br />
a, b. LetVx V f a, x, px p f a, x, nx n f a, x, and let<br />
Va pa na 0. Show that we have:<br />
(a) Vx px nx.
Proof: Given a partition P on a, x, then we have<br />
n<br />
<br />
k1<br />
|f k | |f k | |f k |<br />
kAP<br />
<br />
kAP<br />
kBP<br />
f k |f k |,<br />
kBP<br />
which implies that (taking supermum)<br />
Vx px nx. *<br />
Remark: <strong>The</strong> existence of px and qx is clear, so we know that (*) holds by<br />
<strong>The</strong>orem 1.15.<br />
(b) 0 px Vx and 0 nx Vx.<br />
Proof: Consider a, x, and since<br />
n<br />
Vx |f k | |f k |,<br />
k1 kAP<br />
we know that 0 px Vx. Similarly for 0 nx Vx.<br />
(c) p and n are increasing on a, b.<br />
Proof: Letx, y in a, b with x y, and consider py px as follows. Since<br />
py f k f k ,<br />
kAP, a,y kAP, a,x<br />
we know that<br />
py px.<br />
That is, p is increasing on a, b. Similarly for n.<br />
(d) fx fa px nx. Part (d) gives an alternative proof of <strong>The</strong>orem 6.13.<br />
Proof: Consider a, x, and since<br />
which implies that<br />
fx fa <br />
k1<br />
n<br />
f k f k f k<br />
kAP kBP<br />
fx fa |f k | f k<br />
kBP kAP<br />
which implies that fx fa px nx.<br />
(e) 2px Vx fx fa, 2nx Vx fx fa.<br />
Proof: By (d) and (a), the statement is obvious.<br />
(f) Every point of continuity of f is also a point of continuity of p and of n.<br />
Proof: By (e) and <strong>The</strong>orem 6.14, the statement is obvious.<br />
Curves<br />
6.8 Let f and g be complex-valued functions defined as follows:<br />
ft e 2it if t 0, 1, gt e 2it if t 0, 2.<br />
(a) Prove that f and g have the same graph but are not equivalent according to defintion
in Section 6.12.<br />
Proof: Sinceft : t 0, 1 gt : t 0, 2 the circle of unit disk, we know<br />
that f and g have the same graph.<br />
If f and g are equivalent, then there is an 1-1 and onto function : 0, 2 0, 1 such<br />
that<br />
ft gt.<br />
That is,<br />
e 2it cos2t i sin 2t e 2it cos2t i sin 2t.<br />
In paticular, 1 : c 0, 1. However,<br />
fc cos2c i sin 2c g1 1<br />
which implies that c Z, a contradiction.<br />
(b) Prove that the length of g is twice that of f.<br />
Proof: Since<br />
and<br />
the length of g <br />
0<br />
2<br />
|g t|dt 4<br />
the length of f <br />
0<br />
1<br />
|f t|dt 2,<br />
we know that the length of g is twice that of f.<br />
6.9 Let f be rectifiable path of length L defined on a, b, and assume that f is not<br />
constant on any subinterval of a, b. Lets denote the arc length function given by<br />
sx f a, x if a x b, sa 0.<br />
(a) Prove that s 1 exists and is continuous on 0, L.<br />
Proof: By<strong>The</strong>orem 6.19, we know that sx is continuous and strictly increasing on<br />
0, L. So, the inverse function s 1 exists since s is an 1-1 and onto function, and by<br />
<strong>The</strong>orem 4.29, we know that s 1 is continuous on 0, L.<br />
(b) Define gt fs 1 t if t 0, L and show that g is equivalent to f. Since<br />
ft gst, the function g is said to provide a representation of the graph of f with arc<br />
length as parameter.<br />
Proof: tisclearby<strong>The</strong>orem 6.20.<br />
6.10 Let f and g be two real-valued continuous functions of bounded variation defined<br />
on a, b, with 0 fx gx for each x in a, b, fa ga, fb gb. Leth be the<br />
complex-valued function defined on the interval a,2b a as follows:<br />
ht t ift, ifa t b<br />
2b t ig2b t, ifb t 2b a.<br />
(a) Show that h describes a rectifiable curve .<br />
Proof: It is clear that h is continuous on a,2b a. Note that t, f and g are of bounded<br />
variation on a, b, so h a,2b a exists. That is, h is rectifiable on a,2b a.<br />
(b) Explain, by means of a sketch, the geometric relationship between f, g, andh.<br />
Solution: <strong>The</strong> reader can give it a draw and see the graph lying on x y plane is a
closed region.<br />
(c) Show that the set of points<br />
S x, y : a x b, fx y gx<br />
in a region in R 2 whose boundary is the curve .<br />
Proof: It can be answered by (b), so we omit it.<br />
(d) Let H be the complex-valued function defined on a,2b a as follows:<br />
Ht t 1 igt ft, ifa t b<br />
2<br />
2b t 1 ig2b t f2b t, ifb t 2b a.<br />
2<br />
Show that H describes a rectifiable curve 0 which is the boundary of the region<br />
S 0 x, y : a x b, fx gx 2y gx fx.<br />
Proof: LetFt 1 gt ft and Gt 1 gt ft defined on a, b. Itis<br />
2 2<br />
clear that Ft and Gt are of bounded variation and continuous on a, b with<br />
0 Fx Gx for each x a, b, Fb Gb 0, and Fb Gb 0. In<br />
addition, we have<br />
Ht t iFt, ifa t b<br />
2b t iG2b t, ifb t 2b a.<br />
So, by preceding (a)-(c), we have prove it.<br />
(e) Show that, S 0 has the x axis as a line of symmetry. (<strong>The</strong> region S 0 is called the<br />
symmetrization of S with respect to x axis.)<br />
Proof: Itisclearsincex, y S 0 x, y S 0 by the fact<br />
fx gx 2y gx fx.<br />
(f) Show that the length of 0 does not exceed the length of .<br />
Proof: By (e), the symmetrization of S with respect to x axis tells that<br />
H a, b H b,2b a. So, it suffices to show that h a,2b a 2 H a, b.<br />
Choosing a partition P 1 x 0 a,...,x n b on a, b such that<br />
2 H a, b 2 H P 1 <br />
n<br />
2 x i x i1 2 1 2 f gx i 1 2 f gx i1 2 1/2<br />
i1<br />
n<br />
<br />
i1<br />
4x i x i1 2 f gx i f gx i1 2 1/2<br />
and note that b a 2b a b, we use this P 1 to produce a partition<br />
P 2 P 1 x n b, x n1 b x n x n1 ,...,x 2n 2b a on a,2b a. <strong>The</strong>n we have<br />
*
2n<br />
h P 2 hx i hx i1 <br />
i1<br />
n<br />
2n<br />
hx i hx i1 hx i hx i1 <br />
i1<br />
in1<br />
n<br />
<br />
i1<br />
n<br />
<br />
i1<br />
x i x i1 2 fx i fx i1 2<br />
2n<br />
1/2<br />
<br />
in1<br />
x i x i1 2 gx i gx i1 2 1/2<br />
x i x i1 2 fx i fx i1 2 1/2<br />
xi x i1 2 gx i gx i1 2 1/2<br />
From (*) and (**), we know that<br />
2 H a, b 2 H P 1 h P 2 ***<br />
which implies that<br />
H a,2b a 2 H a, b h a,2b a.<br />
So, we know that the length of 0 does not exceed the length of .<br />
Remark: Definex i x i1 a i , fx i fx i1 b i ,andgx i gx i1 c i , then we<br />
have<br />
4a i2 b i c i 2 1/2<br />
ai2 b i2 1/2 a i2 c i2 1/2 .<br />
Hence we have the result (***).<br />
Proof: It suffices to square both side. We leave it to the reader.<br />
Absolutely continuous functions<br />
A real-valued function f defined on a, b is said to be absolutely continuous on a, b<br />
if for every 0, there is a 0 such that<br />
n<br />
<br />
k1<br />
|fb k fa k | <br />
for every n disjoint open subintervals a k , b k of a, b, n 1, 2, . . . , the sum of whose<br />
n<br />
lengths k1<br />
b k a k is less than .<br />
Absolutely continuous functions occur in the Lebesgue theory of integration and<br />
differentiation. <strong>The</strong> following exercises give some of their elementary properties.<br />
6.11 Prove that every absolutely continuous function on a, b is continuous and of<br />
bounded variation on a, b.<br />
Proof: Letf be absolutely continuous on a, b. <strong>The</strong>n 0, there is a 0 such that<br />
n<br />
<br />
k1<br />
|fb k fa k | <br />
for every n disjoint open subintervals a k , b k of a, b, n 1, 2, . . . , the sum of whose<br />
n<br />
lengths k1<br />
b k a k is less than . So,as|x y| , wherex, y a, b, wehave<br />
|fx fy| .<br />
That is, f is uniformly continuous on a, b. So,f is continuous on a, b.<br />
n<br />
In addition, given any 1, there exists a 0 such that as k1<br />
b k a k ,<br />
where a k , b k s are disjoint open intervals in a, b, wehave<br />
**
n<br />
|fb k fa k | 1.<br />
k1<br />
For this , and let K be the smallest positive integer such that K/2 b a. So,we<br />
partition a, b into K closed subintervals, i.e.,<br />
P y 0 a, y 1 a /2,....,y K1 a K 1/2, y K b. So, it is clear that f is<br />
of bounded variation y i , y i1 ,wherei 0, 1, . . . , K. It implies that f is of bounded<br />
variation on a, b.<br />
Note: <strong>The</strong>re exists functions which are continuous and of bounded variation but<br />
not absolutely continuous.<br />
Remark: 1. <strong>The</strong> standard example is called Cantor-Lebesgue function. <strong>The</strong> reader<br />
can see this in the book, Measure and Integral, An Introduction to <strong>Real</strong> Analysis by<br />
Richard L. Wheeden and Antoni Zygmund, pp 35 and pp 115.<br />
2. If we wrtie ”absolutely continuous” by ABC, ”continuous” by C, and ”bounded<br />
variation” by B, then it is clear that by preceding result, ABC implies B and C, andB and<br />
C do NOT imply ABC.<br />
6.12 Prove that f is absolutely continuous if it satisfies a uniform Lipschitz condition<br />
of order 1 on a, b. (See Exercise 6.2)<br />
Proof: Letf satisfy a uniform Lipschitz condition of order 1 on a, b, i.e.,<br />
|fx fy| M|x y| where x, y a, b. <strong>The</strong>ngiven 0, there is a /M such that<br />
n<br />
as k1<br />
b k a k , wherea k , b k s are disjoint open subintervals on a, b, k 1, . . , n,<br />
we have<br />
n<br />
<br />
k1<br />
|fb k fa k | M|b k a k |<br />
n<br />
k1<br />
n<br />
<br />
k1<br />
Mb k a k <br />
M<br />
.<br />
Hence, f is absolutely continuous on a, b.<br />
6.13 If f and g are absolutely continunous on a, b, prove that each of the following<br />
is also: |f|, cf (c constant), f g, f g; alsof/g if g is bounded away from zero.<br />
Proof: (1) (|f| is absolutely continuous on a, b): Given 0, we want to find a<br />
n<br />
0, such that as k1<br />
b k a k , wherea k , b k s are disjoint open intervals on<br />
a, b, wehave<br />
n<br />
||fb k | |fa k || . 1*<br />
k1<br />
Since f is absolutely continunous on a, b, for this , there is a 0 such that as<br />
n<br />
k1<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave<br />
n<br />
|fb k fa k | <br />
k1<br />
which implies that (1*) holds by the following
n<br />
<br />
k1<br />
n<br />
||fb k | |fa k || |fb k fa k | .<br />
So, we know that |f| is absolutely continuous on a, b.<br />
(2) (cf is absolutely continuous on a, b): If c 0, it is clear. So, we may assume that<br />
n<br />
c 0. Given 0, we want to find a 0, such that as k1<br />
b k a k , where<br />
a k , b k s are disjoint open intervals on a, b, wehave<br />
n<br />
<br />
k1<br />
k1<br />
|cfb k cfa k | . 2*<br />
Since f is absolutely continunous on a, b, for this , there is a 0 such that as<br />
n<br />
k1<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave<br />
n<br />
|fb k fa k | /|c|<br />
k1<br />
which implies that (2*) holds by the following<br />
n<br />
<br />
k1<br />
|cfb k cfa k | |c| |fb k fa k | .<br />
So, we know that cf is absolutely continuous on a, b.<br />
(3) (f g is absolutely continuous on a, b): Given 0, we want to find a 0,<br />
n<br />
such that as k1<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, we<br />
have<br />
n<br />
|f gb k f ga k | . 3*<br />
k1<br />
Since f and g are absolutely continunous on a, b, for this , there is a 0 such that as<br />
n<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave<br />
k1<br />
n<br />
<br />
k1<br />
|fb k fa k | /2 and |gb k ga k | /2<br />
which implies that (3*) holds by the following<br />
n<br />
<br />
k1<br />
n<br />
n<br />
k1<br />
n<br />
k1<br />
|f gb k f ga k |<br />
|fb k fa k gb k ga k |<br />
k1<br />
n<br />
<br />
k1<br />
n<br />
|fb k fa k | <br />
k1<br />
|gb k ga k |<br />
.<br />
So, we know that f g is absolutely continuous on a, b.<br />
(4) (f g is absolutely continuous on a, b.): Let M f sup xa,b |fx| and<br />
M g sup xa,b |gx|. Given 0, we want to find a 0, such that as<br />
n<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave<br />
k1
n<br />
|f gb k f ga k | . 4*<br />
k1<br />
Since f and g are absolutely continunous on a, b, for this , there is a 0 such that as<br />
n<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave<br />
k1<br />
n<br />
n<br />
|fb k fa k | <br />
2M g 1 and |gb k ga k | <br />
k1<br />
k1<br />
which implies that (4*) holds by the following<br />
n<br />
<br />
k1<br />
n<br />
<br />
k1<br />
M f <br />
k1<br />
<br />
.<br />
|f gb k f ga k |<br />
|fb k gb k ga k ga k fb k fa k |<br />
n<br />
n<br />
|gb k ga k | M g <br />
k1<br />
M f<br />
2M f 1 <br />
M g<br />
2M g 1<br />
|fb k fa k |<br />
<br />
2M f 1<br />
Remark: <strong>The</strong> part shows that f n is absolutely continuous on a, b, wheren N, iff is<br />
absolutely continuous on a, b.<br />
(5) (f/g is absolutely continuous on a, b): By (4) it suffices to show that 1/g is<br />
absolutely continuous on a, b. Sinceg is bounded away from zero, say 0 m gx for<br />
n<br />
all x a, b. Given 0, we want to find a 0, such that as k1<br />
b k a k ,<br />
where a k , b k s are disjoint open intervals on a, b, wehave<br />
n<br />
<br />
k1<br />
|1/gb k 1/ga k | . 5*<br />
Since g is absolutely continunous on a, b, for this , there is a 0 such that as<br />
n<br />
k1<br />
b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave<br />
n<br />
|gb k ga k | m 2 <br />
k1<br />
which implies that (4*) holds by the following<br />
n<br />
<br />
k1<br />
n<br />
<br />
k1<br />
|1/gb k 1/ga k |<br />
gb k ga k <br />
gb k ga k <br />
n<br />
1 |gb m 2 k ga k |<br />
k1<br />
.
Supplement on lim sup and lim inf<br />
Introduction<br />
In order to make us understand the information more on approaches of a given real<br />
sequence a n <br />
n1<br />
, we give two definitions, thier names are upper limit and lower limit. It<br />
is fundamental but important tools in analysis. We do NOT give them proofs. <strong>The</strong> reader<br />
can see the book, Infinite Series by Chao Wen-Min, pp 84-103. (Chinese Version)<br />
Definition<br />
Definition of limit sup and limit inf<br />
and<br />
Example<br />
Example<br />
Example<br />
Given a real sequence a n <br />
n1<br />
,wedefine<br />
b n supa m : m ≥ n<br />
1 −1 n <br />
n1<br />
−1 n <br />
n n1<br />
<br />
−n n1<br />
c n infa m : m ≥ n.<br />
0,2,0,2,..., sowehave<br />
b n 2andc n 0 for all n.<br />
−1, 2, −3,4,..., sowehave<br />
b n and c n − for all n.<br />
−1, −2, −3, . . . , sowehave<br />
b n −n and c n − for all n.<br />
Proposition Given a real sequence a n <br />
n1<br />
, and thus define b n and c n as the same as<br />
before.<br />
1 b n ≠−,andc n ≠∀n ∈ N.<br />
2 If there is a positive integer p such that b p , then b n ∀n ∈ N.<br />
If there is a positive integer q such that c q −, then c n − ∀n ∈ N.<br />
3 b n is decreasing and c n is increasing.<br />
By property 3, we can give definitions on the upper limit and the lower limit of a given<br />
sequence as follows.<br />
Definition Given a real sequence a n and let b n and c n as the same as before.<br />
(1) If every b n ∈ R, then<br />
infb n : n ∈ N<br />
is called the upper limit of a n , denoted by<br />
lim n→<br />
sup a n .<br />
That is,<br />
lim n→<br />
sup a n inf b n.<br />
n<br />
If every b n , then we define<br />
lim n→<br />
sup a n .<br />
(2) If every c n ∈ R, then<br />
supc n : n ∈ N<br />
is called the lower limit of a n , denoted by
lim n→<br />
inf a n .<br />
That is,<br />
lim n→<br />
inf a n sup c n .<br />
n<br />
If every c n −, then we define<br />
lim n→<br />
inf a n −.<br />
Remark <strong>The</strong> concept of lower limit and upper limit first appear in the book (Analyse<br />
Alge’brique) written by Cauchy in 1821. But until 1882, Paul du Bois-Reymond<br />
gave explanations on them, it becomes well-known.<br />
Example 1 −1 n <br />
n1<br />
0,2,0,2,..., sowehave<br />
b n 2andc n 0 for all n<br />
which implies that<br />
lim sup a n 2 and lim inf a n 0.<br />
Example −1 n <br />
n n1<br />
−1, 2, −3,4,..., sowehave<br />
b n and c n − for all n<br />
which implies that<br />
lim sup a n and lim inf a n −.<br />
<br />
−n n1<br />
Example −1, −2, −3, . . . , sowehave<br />
b n −n and c n − for all n<br />
which implies that<br />
lim sup a n − and lim inf a n −.<br />
Relations with convergence and divergence for upper (lower) limit<br />
<strong>The</strong>orem Let a n be a real sequence, then a n converges if, and only if, the upper<br />
limit and the lower limit are real with<br />
lim n→<br />
sup a n lim n→<br />
inf a n lim n→<br />
a n .<br />
<strong>The</strong>orem Let a n be a real sequence, then we have<br />
(1) lim n→ sup a n a n has no upper bound.<br />
(2) lim n→ sup a n − for any M 0, there is a positive integer n 0 such<br />
that as n ≥ n 0 ,wehave<br />
a n ≤−M.<br />
(3) lim n→ sup a n a if, and only if, (a) given any 0, there are infinite<br />
many numbers n such that<br />
a − a n<br />
and (b) given any 0, there is a positive integer n 0 such that as n ≥ n 0 ,wehave<br />
a n a .<br />
Similarly, we also have<br />
<strong>The</strong>orem Let a n be a real sequence, then we have
(1) lim n→ inf a n − a n has no lower bound.<br />
(2) lim n→ inf a n for any M 0, there is a positive integer n 0 such<br />
that as n ≥ n 0 ,wehave<br />
a n ≥ M.<br />
(3) lim n→ inf a n a if, and only if, (a) given any 0, there are infinite<br />
many numbers n such that<br />
a a n<br />
and (b) given any 0, there is a positive integer n 0 such that as n ≥ n 0 ,wehave<br />
a n a − .<br />
From <strong>The</strong>orem 2 an <strong>The</strong>orem 3, the sequence is divergent, we give the following<br />
definitios.<br />
Definition Let a n be a real sequence, then we have<br />
(1) If lim n→ sup a n −, then we call the sequence a n diverges to −,<br />
denoted by<br />
lim n→<br />
a n −.<br />
<strong>The</strong>orem<br />
<strong>The</strong>orem<br />
<strong>The</strong>orem<br />
(2) If lim n→ inf a n , then we call the sequence a n diverges to ,<br />
denoted by<br />
lim n→<br />
a n .<br />
Let a n be a real sequence. If a is a limit point of a n , then we have<br />
lim n→<br />
inf a n ≤ a ≤ lim n→<br />
sup a n .<br />
Some useful results<br />
Let a n be a real sequence, then<br />
(1) lim n→ inf a n ≤ lim n→ sup a n .<br />
(2) lim n→ inf−a n − lim n→ sup a n and lim n→ sup−a n − lim n→ inf a n<br />
(3) If every a n 0, and 0 lim n→ inf a n ≤ lim n→ sup a n , then we<br />
have<br />
lim n→<br />
sup a 1 n<br />
1<br />
lim n→ inf a n<br />
Let a n and b n be two real sequences.<br />
(1) If there is a positive integer n 0 such that a n ≤ b n , then we have<br />
lim n→<br />
inf a n ≤ lim n→<br />
inf b n and lim n→<br />
sup a n ≤ lim n→<br />
sup b n .<br />
(2) Suppose that − lim n→ inf a n , lim n→ inf b n , lim n→ sup a n ,<br />
lim n→ sup b n , then<br />
lim n→<br />
inf a n lim n→<br />
inf b n<br />
and lim n→<br />
inf 1 a n<br />
1<br />
lim n→ sup a n<br />
.<br />
≤ lim n→<br />
infa n b n <br />
≤ lim n→<br />
inf a n lim n→<br />
sup b n (or lim n→<br />
sup a n lim n→<br />
inf b n )<br />
≤ lim n→<br />
supa n b n <br />
≤ lim n→<br />
sup a n lim n→<br />
sup b n .
In particular, if a n converges, we have<br />
lim n→<br />
supa n b n lim n→<br />
a n lim n→<br />
sup b n<br />
and<br />
lim n→<br />
infa n b n lim n→<br />
a n lim n→<br />
inf b n .<br />
(3) Suppose that − lim n→ inf a n , lim n→ inf b n , lim n→ sup a n ,<br />
lim n→ sup b n , anda n 0, b n 0 ∀n, then<br />
lim n→<br />
inf a n<br />
lim n→<br />
inf b n<br />
≤ lim n→<br />
infa n b n <br />
≤ lim n→<br />
inf a n lim n→<br />
sup b n (or lim n→<br />
inf b n<br />
≤ lim n→<br />
supa n b n <br />
≤ lim n→<br />
sup a n lim n→<br />
sup b n .<br />
In particular, if a n converges, we have<br />
and<br />
lim n→<br />
supa n b n <br />
lim n→<br />
infa n b n <br />
lim n→<br />
a n<br />
lim n→<br />
a n<br />
lim n→<br />
sup b n<br />
lim n→<br />
inf b n .<br />
lim n→<br />
sup a n )<br />
<strong>The</strong>orem Let a n be a positive real sequence, then<br />
lim n→<br />
inf a n1<br />
a n<br />
≤ lim n→<br />
infa n 1/n ≤ lim n→<br />
supa n 1/n ≤ lim n→<br />
sup a n1<br />
a n<br />
.<br />
Remark We can use the inequalities to show<br />
n! 1/n<br />
lim n→ n 1/e.<br />
<strong>The</strong>orem Let a n be a real sequence, then<br />
lim n→<br />
inf a n ≤ lim n→<br />
inf a 1 ...a n<br />
n<br />
≤ lim n→<br />
sup a 1 ...a n<br />
n<br />
≤ lim n→<br />
sup a n .<br />
Exercise Let f : a, d → R be a continuous function, and a n is a real sequence. If f is<br />
increasing and for every n, lim n→ inf a n , lim n→ sup a n ∈ a, d, then<br />
lim n→<br />
sup fa n f lim n→<br />
sup a n<br />
and lim n→<br />
inf fa n f lim n→<br />
inf a n .<br />
Remark: (1) <strong>The</strong> condition that f is increasing cannot be removed. For<br />
example,<br />
fx |x|,<br />
and<br />
1/k if k is even<br />
a k <br />
−1 − 1/k if k is odd.<br />
(2) <strong>The</strong> proof is easy if we list the definition of limit sup and limit inf. So, we<br />
omit it.<br />
Exercise Let a n be a real sequence satisfying a np ≤ a n a p for all n, p. Show that<br />
an<br />
n converges.<br />
Hint: Consider its limit inf.
Remark: <strong>The</strong> exercise is useful in the theory of Topological Entorpy.<br />
Infinite Series <strong>And</strong> Infinite Products<br />
Sequences<br />
8.1 (a) Given a real-valed sequence a n bounded above, let u n supa k : k ≥ n.<br />
<strong>The</strong>n u n ↘ and hence U lim n→ u n is either finite or −. Prove that<br />
U lim n→<br />
sup a n lim n→<br />
supa k : k ≥ n.<br />
Proof: It is clear that u n ↘ and hence U lim n→ u n is either finite or −.<br />
If U −, then given any M 0, there exists a positive integer N such that as n ≥ N,<br />
we have<br />
u n ≤−M<br />
which implies that, as n ≥ N, a n ≤−M. So, lim n→ a n −. Thatis,a n is not bounded<br />
below. In addition, if a n has a finite limit supreior, say a. <strong>The</strong>ngiven 0, and given<br />
m 0, there exists an integer n m such that<br />
a n a − <br />
which contradicts to lim n→ a n −. From above results, we obtain<br />
U lim n→<br />
sup a n<br />
in the case of U −.<br />
If U is finite, then given 0, there exists a positive integer N such that as n ≥ N, we<br />
have<br />
U ≤ u n U .<br />
So, as n ≥ N, u n U which implies that, as n ≥ N, a n U . In addition, given<br />
′ 0, and m 0, there exists an integer n m,<br />
U − ′ a n<br />
by U ≤ u n supa k : k ≥ n if n ≥ N. From above results, we obtain<br />
U lim n→<br />
sup a n<br />
in the case of U is finite.<br />
(b)Similarly, if a n is bounded below, prove that<br />
V lim n→<br />
inf a n lim n→<br />
infa k : k ≥ n.<br />
Proof: Since the proof is similar to (a), we omit it.<br />
If U and V are finite, show that:<br />
(c) <strong>The</strong>re exists a subsequence of a n which converges to U and a subsequence which<br />
converges to V.<br />
Proof: SinceU lim sup n→ a n by (a), then<br />
(i) Given 0, there exists a positive integer N such that as n ≥ N, wehave<br />
a n U .<br />
(ii) Given 0, and m 0, there exists an integer Pm m,<br />
U − a Pm .<br />
Hence, a Pm is a convergent subsequence of a n with limit U.<br />
Similarly for the case of V.
(d) If U V, every subsequnce of a n converges to U.<br />
Proof: By(a)and(b),given 0, then there exists a positive integer N 1 such that as<br />
n ≥ N 1 ,wehave<br />
a n U <br />
and there exists a positive integer N 2 such that as n ≥ N 2 ,wehave<br />
U − a n .<br />
Hence, as n ≥ maxN 1 , N 2 ,wehave<br />
U − a n U .<br />
That is, a n is a convergent sequence with limit U. So, every subsequnce of a n <br />
converges to U.<br />
8.2 Given two real-valed sequence a n and b n bounded below. Prove hat<br />
(a) lim sup n→ a n b n ≤ lim sup n→ a n lim sup n→ b n .<br />
Proof: Note that a n and b n bounded below, we have lim sup n→ a n or is<br />
finite. <strong>And</strong> lim sup n→ b n or is finite. It is clear if one of these limit superior is ,<br />
sowemayassumethatbotharefinite.Leta lim sup n→ a n and b lim sup n→ b n . <strong>The</strong>n<br />
given 0, there exists a positive integer N such that as n ≥ N, wehave<br />
a n b n a b /2. *<br />
In addition, let c lim sup n→ a n b n ,wherec by (*). So, for the same 0, and<br />
given m N there exists a positive integer K such that as K ≥ N, wehave<br />
c − /2 a K b K . **<br />
By (*) and (**), we obtain that<br />
c − /2 a K b K a b /2<br />
which implies that<br />
c ≤ a b<br />
since is arbitrary. So,<br />
lim sup a n b n ≤ lim sup a n lim sup b n.<br />
n→ n→ n→<br />
Remark: (1) <strong>The</strong> equality may NOT hold. For example,<br />
a n −1 n and b n −1 n1 .<br />
(2) <strong>The</strong> reader should noted that the finitely many terms does NOT change the relation<br />
of order. <strong>The</strong> fact is based on process of proof.<br />
(b) lim sup n→ a n b n ≤ lim sup n→ a n lim sup n→ b n if a n 0, b n 0 for all n, and<br />
if both lim sup n→ a n and lim sup n→ b n are finite or both are infinite.<br />
Proof: Let lim sup n→ a n a and lim sup n→ b n b. It is clear that we may assume<br />
that a and b are finite. Given 0, there exists a positive integer N such that as n ≥ N,<br />
we have<br />
a n b n a b ab a b . *<br />
In addition, let c lim sup n→ a n b n ,wherec by (*). So, for the same 0, and<br />
given m N there exists a positive integer K such that as K ≥ N, wehave<br />
c − a K b K . **<br />
By (*) and (**), we obtain that<br />
c − a K b K a b a b
which implies that<br />
c ≤ a b<br />
since is arbitrary. So,<br />
lim sup a nb n ≤ lim sup a n lim sup b n .<br />
n→ n→ n→<br />
Remark: (1) <strong>The</strong> equality may NOT hold. For example,<br />
a n 1/n if n is odd and a n 1ifn is even.<br />
and<br />
b n 1ifn is odd and b n 1/n if n is even.<br />
(2) <strong>The</strong> reader should noted that the finitely many terms does NOT change the relation<br />
of order. <strong>The</strong> fact is based on the process of the proof.<br />
(3) <strong>The</strong> reader should be noted that if letting A n log a n and B n log b n , thenby(a)<br />
and log x is an increasing function on 0, , we have proved (b).<br />
8.3 Prove that <strong>The</strong>orem 8.3 and 8.4.<br />
(<strong>The</strong>orem 8.3) Leta n be a sequence of real numbers. <strong>The</strong>n we have:<br />
(a) lim inf n→ a n ≤ lim sup n→ a n .<br />
Proof: If lim sup n→ a n , then it is clear. We may assume that<br />
lim sup n→ a n . Hence, a n is bounded above. We consider two cases: (i)<br />
lim sup n→ a n a, wherea is finite and (ii) lim sup n→ a n −.<br />
For case (i), if lim inf n→ a n −, then there is nothing to prove it. We may assume<br />
that lim inf n→ a n a ′ ,wherea ′ is finite. By definition of limit superior and limit inferior,<br />
given 0, there exists a positive integer N such that as n ≥ N, wehave<br />
a ′ − /2 a n a /2<br />
which implies that a ′ ≤ a since is arbitrary.<br />
For case (ii), since lim sup n→ a n −, wehavea n is not bounded below. If<br />
lim inf n→ a n −, then there is nothing to prove it. We may assume that<br />
lim inf n→ a n a ′ ,wherea ′ is finite. By definition of limit inferior, given 0, there<br />
exists a positive integer N such that as n ≥ N, wehave<br />
a ′ − /2 a n<br />
which contradicts that a n is not bounded below.<br />
So, from above results, we have proved it.<br />
(b) <strong>The</strong> sequence converges if and only if, lim sup n→ a n and lim inf n→ a n are both<br />
finite and equal, in which case lim n→ a n lim inf n→ a n lim sup n→ a n .<br />
Proof: ()Given a n a convergent sequence with limit a. So, given 0, there<br />
exists a positive integer N such that as n ≥ N, wehave<br />
a − a n a .<br />
By definition of limit superior and limit inferior, a lim inf n→ a n lim sup n→ a n .<br />
()By definition of limit superior, given 0, there exists a positive integer N 1 such<br />
that as n ≥ N 1 ,wehave<br />
a n a <br />
and by definition of limit superior, given 0, there exists a positive integer N 2 such that<br />
as n ≥ N 2 ,wehave
a − a n .<br />
So, as n ≥ maxN 1 , N 2 ,wehave<br />
a − a n a .<br />
That is, lim n→ a n a.<br />
(c) <strong>The</strong> sequence diverges to if and only if, lim inf n→ a n lim sup n→ a n .<br />
Proof: ()Given a sequence a n with lim n→ a n . So, given M 0, there is a<br />
positive integer N such that as n ≥ N, wehave<br />
M ≤ a n . *<br />
It implies that a n is not bounded above. So, lim sup n→ a n . In order to show that<br />
lim inf n→ a n . We first note that a n is bounded below. Hence, lim inf n→ a n ≠−.<br />
So, it suffices to consider that lim inf n→ a n is not finite. (So, we have<br />
lim inf n→ a n . ). Assume that lim inf n→ a n a, wherea is finite. <strong>The</strong>n given 1,<br />
and an integer m, there exists a positive Km m such that<br />
a Km a 1<br />
which contradicts to (*) if we choose M a 1. So, lim inf n→ a n is not finite.<br />
(d) <strong>The</strong> sequence diverges to − if and only if, lim inf n→ a n lim sup n→ a n −.<br />
Proof: Note that, lim sup n→ −a n − lim inf n→ a n . So, by (c), we have proved it.<br />
(<strong>The</strong>orem 8.4)Assume that a n ≤ b n for each n 1,2,.... <strong>The</strong>n we have:<br />
lim n→<br />
inf a n ≤ lim n→<br />
inf b n and lim n→<br />
sup a n ≤ lim n→<br />
sup b n .<br />
Proof: If lim inf n→ b n , there is nothing to prove it. So, we may assume that<br />
lim inf n→ b n . That is, lim inf n→ b n − or b, whereb is finite.<br />
For the case, lim inf n→ b n −, it means that the sequence a n is not bounded<br />
below. So, b n is also not bounded below. Hence, we also have lim inf n→ a n −.<br />
For the case, lim inf n→ b n b, whereb is finite. We consider three cases as follows.<br />
(i) if lim inf n→ a n −, then there is nothing to prove it.<br />
(ii) if lim inf n→ a n a, wherea is finite. Given 0, then there exists a positive<br />
integer N such that as n ≥ N<br />
a − /2 a n ≤ b n b /2<br />
which implies that a ≤ b since is arbitrary.<br />
(iii) if lim inf n→ a n , then by <strong>The</strong>orem 8.3 (a) and (c), we know that<br />
lim n→ a n which implies that lim n→ b n . Also, by <strong>The</strong>orem 8.3 (c), wehave<br />
lim inf n→ b n which is absurb.<br />
So, by above results, we have proved that lim inf n→ a n ≤ lim inf n→ b n .<br />
Similarly, we have lim sup n→ a n ≤ lim sup n→ b n .<br />
8.4 If each a n 0, prove that<br />
lim n→<br />
inf a n1<br />
a n<br />
≤ lim n→<br />
infa n 1/n ≤ lim n→<br />
supa n 1/n ≤ lim n→<br />
sup a n1<br />
a n<br />
.<br />
Proof: By<strong>The</strong>orem 8.3 (a), it suffices to show that<br />
lim n→<br />
inf a n1<br />
a n<br />
We first prove<br />
≤ lim n→<br />
infa n 1/n and lim n→<br />
supa n 1/n ≤ lim n→<br />
sup a n1<br />
a n<br />
.<br />
lim n→<br />
supa n 1/n ≤ lim n→<br />
sup a n1<br />
a n<br />
.
If lim sup n→<br />
a n1<br />
a n<br />
, then it is clear. In addition, since a n1<br />
a n<br />
is positive,<br />
a<br />
≠−. So, we may assume that lim sup n1<br />
n→ a n<br />
a, wherea is finite.<br />
Given 0, then there exists a positive integer N such that as n ≥ N, wehave<br />
a n1<br />
a n<br />
a <br />
lim sup n→<br />
a n1<br />
a n<br />
which implies that<br />
So,<br />
which implies that<br />
a Nk a N a k ,wherek 1,2,....<br />
a Nk 1<br />
Nk<br />
a N 1<br />
Nk a <br />
k<br />
Nk<br />
lim supa Nk 1<br />
Nk<br />
k→<br />
≤ lim supa N 1<br />
k<br />
Nk a <br />
Nk<br />
k→<br />
a .<br />
So,<br />
lim supa Nk 1<br />
Nk<br />
≤ a<br />
k→<br />
since is arbitrary. Note that the finitely many terms do NOT change the value of limit<br />
superiror of a given sequence. So, we finally have<br />
lim n→<br />
supa n 1/n ≤ a lim n→<br />
sup a n1<br />
a n<br />
.<br />
Similarly for<br />
lim n→<br />
inf a n1<br />
a n<br />
≤ lim n→<br />
infa n 1/n .<br />
Remark: <strong>The</strong>se ineqaulities is much important; we suggest that the reader keep it mind.<br />
At the same time, these inequalities tells us that the root test is more powerful than the<br />
ratio test. We give an example to say this point. Given a series<br />
1<br />
2 1 3 1 2 1 2 3 ... 1 2 2 n 3 1 n ...<br />
where<br />
n n<br />
a 2n−1 1 ,anda2n 1 , n 1, 2, . . .<br />
2 3<br />
with<br />
and<br />
lim n→<br />
inf a n1<br />
a n<br />
lim n→<br />
supa n 1/n 1<br />
2<br />
1<br />
0, lim n→<br />
sup a n1<br />
a n<br />
.<br />
8.5 Let a n n n /n!. Show that lim n→ a n1 /a n e and use Exercise 8.4 to deduce that<br />
lim n<br />
n→<br />
e.<br />
n! 1/n<br />
Proof: Since<br />
a n1<br />
a n<br />
n 1n1 n!<br />
n 1!n n 1 1 n<br />
n → e,<br />
by Exercise 8.4, wehave<br />
lim n→<br />
a n 1/n lim n<br />
n→<br />
e.<br />
n! 1/n
Remark: <strong>The</strong>re are many methods to show this. We do NOT give the detailed proof.<br />
But there are hints.<br />
(1) Taking log on n!<br />
n n 1/n , and thus consider<br />
1<br />
n log 1 n ... log n n → <br />
0<br />
1<br />
log xdx −1.<br />
(2) Stirling’s Formula:<br />
n! n n e −n 2n e <br />
12n ,where ∈ 0, 1.<br />
Note: In general, we have<br />
lim<br />
x→<br />
Γx 1<br />
x x e −x 2x 1,<br />
where Γx is the Gamma Function. <strong>The</strong> reader can see the book, Principles of<br />
Mathematical Analysis by Walter Rudin, pp 192-195.<br />
(3) Note that 1 1 x x ↗ e and 1 1 x x1 ↘ e on 0, . So,<br />
1 1 n n1<br />
n e 1 1 n<br />
which implies that<br />
en n e −n n! en n1 e −n .<br />
(4) Using O-Stolz’s <strong>The</strong>orem: Let lim n→ y n and y n ↗.If<br />
lim<br />
x n1 − x n<br />
n→ y n1 − y n<br />
a, wherea is finite or ,<br />
then<br />
lim<br />
x n<br />
n→ y n<br />
a.<br />
Let x n log 1 n ... log n n and y n n.<br />
Note: For the proof of O-Stolz’s <strong>The</strong>orem, the reader can see the book, An<br />
Introduction to Mathematical Analysis by Loo-Keng Hua, pp 195. (Chinese Version)<br />
(5) Note that, if a n is a positive sequence with lim n→ a n a, then<br />
a 1 a n 1/n → a as n → .<br />
Taking a n 1 1 n n , then<br />
a 1 a n 1/n n n<br />
1 <br />
n!<br />
1 n → e.<br />
Note: For the proof, it is easy from the Exercise 8.6. We give it a proof as follows. Say<br />
lim n→ a n a. Ifa 0, then by A. P. ≥ G. P. , we have<br />
a 1 a n 1/n ≤ a 1 ...a n<br />
n → 0byExercise 8.6.<br />
So, we consider a ≠ 0 as follows. Note that log a n → log a. So, by Exercise 8.6,<br />
log a 1 ... log a n<br />
n → log a<br />
which implies that a 1 a n 1/n → a.<br />
8.6 Let a n be real-valued sequence and let n a 1 ...a n /n. Show that<br />
lim n→<br />
inf a n ≤ lim n→<br />
inf n ≤ lim n→<br />
sup n ≤ lim n→<br />
sup a n .<br />
Proof: By<strong>The</strong>orem 8.3 (a), it suffices to show that<br />
lim n→<br />
inf a n ≤ lim n→<br />
inf n and lim n→<br />
sup n ≤ lim n→<br />
sup a n .<br />
1/n
We first prove<br />
lim n→<br />
sup n ≤ lim n→<br />
sup a n .<br />
If lim sup n→ a n , there is nothing to prove it. We may assume that<br />
lim sup n→ a n − or a, wherea is finite.<br />
For the case, lim sup n→ a n −, then by <strong>The</strong>orem 8.3 (d), wehave<br />
lim n→<br />
a n −.<br />
So, given M 0, there exists a positive integer N such that as n ≥ N, wehave<br />
a n ≤−M. *<br />
Let n N, wehave<br />
n a 1 ...a N ..a n<br />
n<br />
a 1 ...a N<br />
n<br />
a N1 ...a n<br />
n<br />
≤ a 1 ...a N<br />
n n − N<br />
n −M<br />
which implies that<br />
lim n→<br />
sup n ≤−M.<br />
Since M is arbitrary, we finally have<br />
lim n→<br />
sup n −.<br />
For the case, lim sup n→ a n a, wherea is finite. Given 0, there exists a positive<br />
integer N such that as n ≥ N, wehave<br />
a n a .<br />
Let n N, wehave<br />
n a 1 ...a N ..a n<br />
n<br />
a 1 ...a N<br />
n<br />
a N1 ...a n<br />
n<br />
≤ a 1 ...a N<br />
n n − N<br />
n<br />
a <br />
which implies that<br />
lim n→<br />
sup n ≤ a <br />
which implies that<br />
lim n→<br />
sup n ≤ a<br />
since is arbitrary.<br />
Hence, from above results, we have proved that lim sup n→ n ≤ lim sup n→ a n .<br />
Similarly for lim inf n→ a n ≤ lim inf n→ n .<br />
Remark: We suggest that the reader keep it in mind since it is the fundamental and<br />
useful in the theory of Fourier Series.<br />
8.7 Find lim sup n→ a n and lim inf n→ a n if a n is given by<br />
(a) cosn<br />
Proof: Note that, a b : a, b ∈ Z is dense in R. Bycosn cosn 2k, we<br />
know that<br />
lim n→<br />
sup cos n 1 and lim n→<br />
inf cos n −1.
Remark: <strong>The</strong> reader may give it a try to show that<br />
lim n→<br />
sup sin n 1 and lim n→<br />
inf sin n −1.<br />
(b) 1 1 n cosn<br />
Proof: Note that<br />
So, it is clear that<br />
lim n→<br />
sup 1 1 n<br />
1 1 n cosn <br />
1ifn 2k<br />
−1 ifn 2k − 1 .<br />
cosn 1 and lim n→<br />
inf 1 1 n cosn −1.<br />
(c) n sin n 3<br />
Proof: Note that as n 1 6k, n sin n 1 6k sin ,andasn 4 6k,<br />
3 3<br />
n −4 6k sin . So, it is clear that<br />
3<br />
lim n→<br />
sup n sin n and lim<br />
3<br />
inf n sin n n→ 3<br />
−.<br />
(d) sin n cos n 2 2<br />
Proof: Note that sin n cos n 1 sin n 0, we have<br />
2 2 2<br />
lim n→<br />
sup sin n 2 cos n 2<br />
lim inf sin n n→ 2 cos n 2 0.<br />
(e) −1 n n/1 n n<br />
Proof: Note that<br />
we know that<br />
(f) n 3<br />
− n 3 <br />
Proof: Note that<br />
So, it is clear that<br />
lim n→<br />
−1 n n/1 n n 0,<br />
lim n→<br />
sup−1 n n/1 n n lim n→<br />
inf−1 n n/1 n n 0.<br />
n<br />
3 − n 3<br />
<br />
1<br />
3<br />
2<br />
3<br />
if n 3k 1<br />
if n 3k 2<br />
0ifn 3k<br />
,wherek 0, 1, 2, . . . .<br />
lim n→<br />
sup n 3 − n <br />
3<br />
2 and lim<br />
3 inf n n→ 3 − n 3<br />
Note.In(f),x denoted the largest integer ≤ x.<br />
n<br />
8.8 Let a n 2 n − ∑ k1<br />
1/ k . Prove that the sequence a n converges to a limit p<br />
in the interval 1 p 2.<br />
n<br />
n<br />
Proof: Consider ∑ k1<br />
1/ k : S n and x<br />
−1/2<br />
dx : T n , then<br />
1<br />
lim n→<br />
d n exists, where d n S n − T n<br />
by Integral Test. We denote the limit by d, then<br />
0.
0 ≤ d 1 *<br />
by <strong>The</strong>orem 8.23 (i). Note that d n − fn is a positive increasing sequence, so we have<br />
d 0. **<br />
Since<br />
T n 2 n − 2<br />
which implies that<br />
lim n→<br />
n<br />
2 n − ∑ 1/ k lim n→<br />
a n 2 − d p.<br />
k1<br />
By (*) and (**), we have proved that 1 p 1.<br />
Remark: (1) <strong>The</strong> use of Integral Test is very useful since we can know the behavior<br />
of a given series by integral. However, in many cases, the integrand may be so complicated<br />
that it is not easy to calculate. For example: Prove that the convergence of<br />
<br />
∑<br />
n2<br />
1<br />
nlog n p ,wherep 1.<br />
Of course, it can be checked by Integral Test. But there is the <strong>The</strong>orem called Cauchy<br />
Condensation <strong>The</strong>orem much powerful than Integral Test in this sense. In addition, the<br />
reader can think it twice that in fact, Cauchy condensation <strong>The</strong>orem is equivalent to<br />
Integral Test.<br />
(Cauchy Condensation <strong>The</strong>orem)Let a n be a positive decreasing sequence. <strong>The</strong>n<br />
<br />
∑<br />
n1<br />
a n converges if, and only if, ∑<br />
k0<br />
<br />
2 k a 2<br />
k converges.<br />
Note: (1) <strong>The</strong> proof is not hard; the reader can see the book, Principles of<br />
Mathematical Analysis by Walter Rudin, pp 61-63.<br />
(2) <strong>The</strong>re is an extension of Cauchy Condensation <strong>The</strong>orem (Oskar Schlomilch):<br />
Suppose that a k be a positive and decreasing sequence and m k ⊆ N is a sequence. If<br />
there exists a c 0 such that<br />
0 m k2 − m k1 ≤ cm k1 − m k for all k,<br />
then<br />
<br />
∑<br />
k1<br />
<br />
a k converges if, and only if, ∑m k1 − m k a mk .<br />
k0<br />
Note: <strong>The</strong> proof is similar with Cauchy Condensation <strong>The</strong>orem, soweomitit.<br />
(2) <strong>The</strong>re is a similar <strong>The</strong>orem, we write it as a reference. If t ≥ a, ft is a<br />
non-negative increasing function, then as x ≥ a, wehave<br />
x<br />
∑ fn − ftdt<br />
a≤n≤x<br />
a<br />
≤ fx.<br />
Proof: <strong>The</strong> proof is easy by drawing a graph. So, we omit it.<br />
P.S.: <strong>The</strong> theorem is useful when we deal with some sums. For example,<br />
ft log t.<br />
<strong>The</strong>n
∑ log n − x log x x − 1 ≤ log x.<br />
1≤n≤x<br />
In particular, as x ∈ N, we thus have<br />
n log n − n 1 − log n ≤ log n! ≤ n log n − n 1 log n<br />
which implies that<br />
n n−1 e −n1 ≤ n! ≤ n n1 e −n1 .<br />
In each of Exercise 8.9. through 8.14, show that the real-valed sequence a n is<br />
convergent. <strong>The</strong> given conditions are assumed to hold for all n ≥ 1. In Exercise 8.10<br />
through 8.14, show that a n has the limit L indicated.<br />
8.9 |a n| 2, |a n2 − a n1 | ≤ 1 8 |a 2<br />
n1 − a n2 |.<br />
Proof: Since<br />
|a n2 − a n1 | ≤ 1 8 |a 2<br />
n1 − a n2 |<br />
1 8 |a n1 − a n||a n1 a n|<br />
we know that<br />
So,<br />
≤ 1 2 |a n1 − a n| since |a n| 2<br />
|a n1 − a n| ≤ 1<br />
2<br />
n−1<br />
|a2 − a 1 | ≤ 1<br />
2<br />
n−3<br />
.<br />
k<br />
|a nk − a n| ≤ ∑|a nj − a nj−1 |<br />
j1<br />
k<br />
≤ ∑<br />
j1<br />
1<br />
2<br />
nj−4<br />
≤ 1<br />
2<br />
n−2<br />
→ as n → .<br />
Hence, a n is a Cauchy sequence. So, a n is a convergent sequence.<br />
Remark: (1) If |a n1 − a n| ≤ b n for all n ∈ N, and∑ b n converges, then ∑ a n<br />
converges.<br />
Proof: Since the proof is similar with the Exercise, we omit it.<br />
(2) In (1), the condition ∑ b n converges CANNOT omit. For example,<br />
n 1<br />
(i) Let a n sin ∑ k1<br />
Or<br />
k<br />
(ii) a n is defined as follows:<br />
a 1 1, a 2 1/2, a 3 0, a 4 1/4, a 5 1/2, a 6 3/4, a 7 1, and so on.<br />
8.10 a 1 ≥ 0, a 2 ≥ 0, a n2 a n a n1 1/2 , L a 1 a 22 1/3 .<br />
Proof: If one of a 1 or a 2 is 0, then a n 0 for all n ≥ 2. So, we may assume that<br />
a 1 ≠ 0anda 2 ≠ 0. So, we have a n ≠ 0 for all n. Letb n a n1<br />
a n<br />
, then<br />
b n1 1/ b n for all n<br />
which implies that
Consider<br />
which implies that<br />
which implies that<br />
b n1 b 1 −1<br />
2<br />
n<br />
→ 1asn → .<br />
n1 j2 b j n<br />
j1 b j −1/2<br />
a 1 1/2 a 2<br />
−2/3<br />
an1 1<br />
b n1<br />
2/3<br />
lim n→<br />
a n1 a 1 a 22 1/3 .<br />
Remark: <strong>The</strong>re is another proof. We write it as a reference.<br />
Proof: If one of a 1 or a 2 is 0, then a n 0 for all n ≥ 2. So, we may assume that<br />
a 1 ≠ 0anda 2 ≠ 0. So, we have a n ≠ 0 for all n. Leta 2 ≥ a 1 .Sincea n2 a n a n1 1/2 ,<br />
then inductively, we have<br />
a 1 ≤ a 3 ≤...≤ a 2n−1 ≤...≤ a 2n ≤...≤ a 4 ≤ a 2 .<br />
So, both of a 2n and a 2n−1 converge. Say<br />
lim n→<br />
a 2n x and lim n→<br />
a 2n−1 y.<br />
Note that a 1 ≠ 0anda 2 ≠ 0, so x ≠ 0, and y ≠ 0. In addition, x y by<br />
a n2 a n a n1 1/2 . Hence, a n converges to x.<br />
By a n2 a n a n1 1/2 , and thus<br />
n<br />
j1 a2 j2 n<br />
j1 a j a j1 a 1 a 22 a n1 n−2 j1 a2<br />
j2<br />
which implies that<br />
which implies that<br />
a n1 a2 n2 a 1 a2<br />
2<br />
lim n→<br />
a n x a 1 a 22 1/3 .<br />
8.11 a 1 2, a 2 8, a 2n1 1 2 a 2n a 2n−1 , a 2n2 a 2na 2n−1<br />
a 2n1<br />
, L 4.<br />
Proof: First, we note that<br />
a 2n1 a 2n a 2n−1<br />
2<br />
≥ a 2n a 2n−1 by A. P. ≥ G. P. *<br />
for n ∈ N. So, by a 2n2 a 2na 2n−1<br />
a 2n1<br />
and (*),<br />
a 2n2 a 2na 2n−1<br />
a 2n1<br />
≤ a 2n a 2n−1 ≤ a 2n1 for all n ∈ N.<br />
Hence, by Mathematical Induction, itiseasytoshowthat<br />
a 4 ≤ a 6 ≤...≤ a 2n2 ≤...≤ a 2n1 ≤...≤ a 5 ≤ a 3<br />
for all n ∈ N. It implies that both of a 2n and a 2n−1 converge, say<br />
lim n→<br />
a 2n x and lim n→<br />
a 2n−1 y.<br />
With help of a 2n1 1 a 2 2n a 2n−1 , we know that x y. In addition, by a 2n2 a 2na 2n−1<br />
a 2n1<br />
,<br />
a 1 2, and a 2 8, we know that x 4.<br />
8.12 a 1 −3 2 n1 2 a n3 , L 1. Modify a 1 to make L −2.<br />
Proof: ByMathematical Induction, itiseasytoshowthat<br />
− 2 ≤ a n ≤ 1 for all n. *
So,<br />
3a n1 − a n a n3 − 3a n 2 ≥ 0<br />
by (*) and fx x 3 − 3x 2 x − 1 2 x 2 ≥ 0on−2, 1. Hence, a n is an<br />
increasing sequence with a upper bound 1. So, a n is a convergent sequence with limit L.<br />
So,by3a n1 2 a n3 ,<br />
L 3 − 3L 2 0<br />
which implies that<br />
L 1or − 2.<br />
So, L 1sinca n ↗ and a 1 −3/2.<br />
In order to make L −2, it suffices to let a 1 −2, then a n −2 for all n.<br />
8.13 a 1 3, a n1 31an<br />
3a n<br />
, L 3 .<br />
Proof: By Mathematical Induction, itiseasytoshowthat<br />
a n ≥ 3 for all n. *<br />
So,<br />
a n1 − a n 3 − a n 2<br />
≤ 0<br />
3 a n<br />
which implies that a n is a decreasing sequence. So, a n is a convergent sequence with<br />
limit L by (*). Hence,<br />
31 L<br />
L <br />
3 L<br />
which implies that<br />
L 3 .<br />
So, L 3 since a n ≥ 3 for all n.<br />
and<br />
8.14 a n b n1<br />
b n<br />
,whereb 1 b 2 1, b n2 b n b n1 , L 1 5<br />
Hint. Show that b n2 b n − b2 n1 −1 n1 and deduce that |a n − a n1 | n −2 ,ifn 4.<br />
Proof: ByMathematical Induction, itiseasytoshowthat<br />
Thus, (Note that b n ≠ 0 for all n)<br />
|a n1 − a n| b n2<br />
b n1<br />
b n2 b n − b2 n1 −1 n1 for all n<br />
b n ≥ n if n 4<br />
2<br />
.<br />
− b n1<br />
−1n1 ≤ 1<br />
b n b n b n1 nn 1 1 if n 4.<br />
n2 So, a n is a Cauchy sequence. In other words, a n is a convergent sequence, say<br />
lim n→ b n L. <strong>The</strong>n by b n2 b n b n1 ,wehave<br />
b n2<br />
b n<br />
1<br />
b n1 b n1<br />
which implies that (Note that 0 ≠L ≥ 1sincea n ≥ 1 for all n)<br />
L <br />
L 1 1<br />
which implies that<br />
L 1 5<br />
2<br />
.
So, L 1 5 since L ≥ 1.<br />
2<br />
Remark: (1) <strong>The</strong> sequence b n is the famous sequence named Fabonacci sequence.<br />
<strong>The</strong>re are many researches around it. Also, it is related with so called Golden Section,<br />
5 −1<br />
0.618....<br />
2<br />
(2) <strong>The</strong> reader can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s by G.<br />
H. Hardy and E. M. Wright, Chapter X. <strong>The</strong>n it is clear by continued fractions.<br />
(3) <strong>The</strong>re is another proof. We write it as a reference.<br />
Proof: (STUDY) Sinceb n2 b n b n1 , we may think<br />
x n2 x n x n1 ,<br />
and thus consider x 2 x 1. Say and are roots of x 2 x 1, with . <strong>The</strong>nlet<br />
F n n − n<br />
− ,<br />
we have<br />
F n b n .<br />
So,itiseasytoshowthatL 1 5 . We omit the details.<br />
2<br />
Note: <strong>The</strong> reader should be noted that there are many methods to find the formula of<br />
Fabonacci sequence like F n . For example, using the concept of Eigenvalues if we can<br />
find a suitable matrix.<br />
Series<br />
8.15 Test for convergence (p and q denote fixed rela numbers).<br />
(a) ∑ <br />
n1<br />
n 3 e −n<br />
Proof: ByRoot Test, wehave<br />
So, the series converges.<br />
(b) ∑ <br />
n2<br />
log n p<br />
lim n→<br />
sup<br />
n3 1/n<br />
e 1/e 1.<br />
n<br />
Proof: We consider 2 cases: (i) p ≥ 0, and (ii) p 0.<br />
For case (i), the series diverges since log n p does not converge to zero.<br />
For case (ii), the series diverges by Cauchy Condensation <strong>The</strong>orem (or Integral<br />
Test.)<br />
(c) ∑ n1<br />
p n n p (p 0)<br />
Proof: ByRoot Test, wehave<br />
lim n→<br />
sup<br />
pn 1/n<br />
n p.<br />
p<br />
So, as p 1, the series diverges, and as p 1, the series converges. For p 1, it is clear<br />
that the series ∑ n diverges. Hence,<br />
and<br />
<br />
∑<br />
n1<br />
p n n p converges if p ∈ 0, 1
∑ p n n p diverges if p ∈ 1, .<br />
n1<br />
1<br />
(d) ∑ n2 n p −n<br />
(0 q p)<br />
q<br />
1<br />
Proof: Note that<br />
n p −n<br />
1 1<br />
q n<br />
. We consider 2 cases: (i) p 1 and (ii) p ≤ 1.<br />
p 1−n q−p<br />
For case (i), by Limit Comparison Test with 1 n<br />
, p<br />
1<br />
n<br />
lim<br />
p −n q<br />
n→ 1<br />
1,<br />
n p<br />
the series converges.<br />
For case (ii), by Limit Comparison Test with 1 n<br />
, p<br />
the series diverges.<br />
(e) ∑ <br />
n1<br />
n −1−1/n<br />
1<br />
n<br />
lim<br />
p −n q<br />
n→ 1<br />
n p 1,<br />
Proof: Sincen −1−1/n ≥ n −1 for all n, the series diverges.<br />
1<br />
(f) ∑ n1 p n −q<br />
(0 q p)<br />
n<br />
1<br />
Proof: Note that<br />
p n −q<br />
1 1<br />
n p n 1−<br />
q n . We consider 2 cases: (i) p 1 and (ii) p ≤ 1.<br />
p<br />
For case (i), by Limit Comparison Test with 1 p<br />
, n<br />
1<br />
p<br />
lim<br />
n −q n<br />
n→<br />
1,<br />
1<br />
p n<br />
the series converges.<br />
For case (ii), by Limit Comparison Test with 1 p n ,<br />
the series diverges.<br />
1<br />
(g) ∑ n1 nlog11/n<br />
1<br />
p<br />
lim<br />
n −q n<br />
n→<br />
1,<br />
1<br />
p n<br />
Proof: Since<br />
lim 1<br />
n→ n log1 1/n 1,<br />
we know that the series diverges.<br />
1<br />
(h) ∑ n2 logn log n<br />
Proof: Since the identity a logb b loga ,wehave<br />
log n logn nlog logn<br />
So, the series converges.<br />
1<br />
(i) ∑ n3 nlognlog logn p<br />
≥ n 2 as n ≥ n 0 .<br />
Proof: We consider 3 cases: (i) p ≤ 0, (ii) 0 p 1 and (iii) p 1.
For case (i), since<br />
1<br />
n log nlog log n p ≥ 1 for n ≥ 3,<br />
n log n<br />
1<br />
we know that the series diverges by the divergence of ∑ .<br />
n3 n logn<br />
For case (ii), we consider (choose n 0 large enough)<br />
<br />
∑<br />
jn 0<br />
2 j<br />
2 j log 2 j log log 2 j p 1<br />
log 2 ∑ jn 0<br />
<br />
≥ ∑<br />
jn 0<br />
<br />
1<br />
jlog j p ,<br />
1<br />
jlog j log 2 p<br />
then, by Cauchy Condensation <strong>The</strong>orem, the series diverges since ∑ jn0<br />
by using Cauchy Condensation <strong>The</strong>orem again.<br />
For case (iii), we consider (choose n 0 large enough)<br />
<br />
∑<br />
jn 0<br />
2 j<br />
2 j log 2 j log log 2 j p 1<br />
log 2 ∑ jn 0<br />
<br />
≤ 2 ∑<br />
jn 0<br />
<br />
≤ 4 ∑<br />
jn 0<br />
<br />
1<br />
jlog j log 2 p<br />
1<br />
jlog j log 2 p<br />
1<br />
jlog j p ,<br />
1<br />
jlogj p<br />
1<br />
jlogj p<br />
diverges<br />
then, by Cauchy Condensation <strong>The</strong>orem, the series converges since ∑ jn0<br />
converges by using Cauchy Condensation <strong>The</strong>orem again.<br />
Remark: <strong>The</strong>re is another proof by Integral Test. We write it as a reference.<br />
1<br />
Proof: It is easy to check that fx is continous, positive, and<br />
xlogxlog logx p<br />
decreasing to zero on a, where a 0 for each fixed p. Consider<br />
<br />
<br />
<br />
dx<br />
dy<br />
a x log xlog log x p <br />
log loga y p<br />
which implies that the series converges if p 1 and diverges if p ≤ 1byIntegral Test.<br />
1<br />
(j) ∑ n3 log logn<br />
log logn<br />
log logn<br />
1<br />
Proof: Leta n <br />
log logn for n ≥ 3andbn 1/n, then<br />
log logn<br />
a n<br />
n 1<br />
b n log log n<br />
e −ylogy−ey <br />
→.<br />
So, by Limit Comparison Test, the series diverges.<br />
<br />
(k) ∑ n1<br />
1 n 2 − n<br />
Proof: Note that<br />
1 n 2 − n 1<br />
1 n 2 n ≥ 1 for all n.<br />
1 2 n<br />
So, the series diverges.
(l) ∑ n2<br />
n p 1<br />
n−1 − 1 n<br />
Proof: Note that<br />
n p 1<br />
n − 1 −<br />
1 n<br />
1<br />
n 3 2 −p<br />
n<br />
n − 1<br />
1<br />
1 <br />
n−1<br />
n<br />
So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit<br />
Comparison Test.<br />
<br />
(m) ∑ n1<br />
n 1/n − 1 n<br />
Proof: With help of Root Test,<br />
lim n→<br />
sup n 1/n − 1 n 1/n 0 1,<br />
the series converges.<br />
(n) ∑ n1<br />
n p n 1 − 2 n n − 1<br />
Proof: Note that<br />
n p n 1 − 2 n n − 1<br />
1<br />
n 3 2<br />
n 3 2 −p n n 1 n n − 1 n − 1 n 1<br />
.<br />
So, as p 1/2, the series converges and as p ≥ 1/2, the series diverges by Limit<br />
Comparison Test.<br />
8.16 Let S n 1 , n 2 ,... denote the collection of those positive integers that do not<br />
involve the digit 0 is their decimal representation. (For example, 7 ∈ S but 101 ∉ S.)<br />
Show that ∑ k1<br />
1/n k converges and has a sum less than 90.<br />
Proof: DefineS j the j − digit number ⊆ S. <strong>The</strong>n#S j 9 j and S <br />
j1 S j . Note<br />
that<br />
∑ 1/n k 9 j<br />
10 . j−1 k∈S j<br />
So,<br />
<br />
∑<br />
k1<br />
<br />
1/n k ≤ ∑<br />
<br />
j1<br />
9 j<br />
10 j−1 90.<br />
In addition, it is easy to know that ∑ k1<br />
1/n k ≠ 90. Hence, we have proved that ∑ k1<br />
1/n k<br />
converges and has a sum less than 90.<br />
8.17 Given integers a 1 , a 2 ,...such that 1 ≤ a n ≤ n − 1, n 2, 3, . . . Show that the<br />
<br />
sum of the series ∑ n1<br />
a n /n! is rational if and only if there exists an integer N such that<br />
a n n − 1 for all n ≥ N. Hint: For sufficency, show that ∑ <br />
n2<br />
n − 1/n! is a telescoping<br />
series with sum 1.<br />
Proof: ()Assume that there exists an integer N such that a n n − 1 for all n ≥ N.<br />
<strong>The</strong>n<br />
.
()Assume that ∑ <br />
n1<br />
<br />
That is, p! ∑ np1<br />
a n<br />
p! ∑<br />
np1<br />
<br />
∑<br />
n1<br />
an<br />
n!<br />
N−1<br />
∑<br />
n1<br />
N−1<br />
∑<br />
n1<br />
N−1<br />
∑<br />
n1<br />
<br />
an<br />
n! ∑<br />
nN<br />
an<br />
n!<br />
<br />
an<br />
n! ∑ n − 1<br />
n!<br />
nN<br />
<br />
an<br />
n! ∑<br />
nN<br />
1<br />
n − 1! − 1 n!<br />
N−1<br />
∑<br />
an<br />
n! 1<br />
N − 1! ∈ Q.<br />
n1<br />
a n /n! is rational, say q p ,whereg. c. d. p, q 1. <strong>The</strong>n<br />
n!<br />
∈ Z. Note that<br />
<br />
a n<br />
n!<br />
<br />
≤ p! ∑<br />
np1<br />
<br />
p! ∑<br />
n1<br />
n − 1<br />
n!<br />
an<br />
n!<br />
∈ Z.<br />
p!<br />
p! 1since1 ≤ a n ≤ n − 1.<br />
So, a n n − 1 for all n ≥ p 1. That is, there exists an integer N such that a n n − 1for<br />
all n ≥ N.<br />
Remark: From this, we have proved that e is irrational. <strong>The</strong> reader should be noted that<br />
we can use <strong>The</strong>orem 8.16 to show that e is irrational by considering e −1 . Since it is easy,<br />
we omit the proof.<br />
8.18 Let p and q be fixed integers, p ≥ q ≥ 1, and let<br />
pn<br />
x n ∑<br />
kqn1<br />
n<br />
1<br />
k<br />
, s n ∑<br />
k1<br />
−1 k1<br />
k<br />
(a) Use formula (8) to prove that lim n→ x n logp/q.<br />
Proof: Since<br />
we know that<br />
n<br />
∑<br />
k1<br />
1<br />
k<br />
log n r O 1 n ,<br />
pn<br />
x n ∑<br />
k1<br />
qn<br />
1 − k<br />
∑<br />
k1<br />
logp/q O 1 n<br />
which implies that lim n→ x n logp/q.<br />
(b) When q 1, p 2, show that s 2n x n and deduce that<br />
<br />
∑<br />
n1<br />
1<br />
k<br />
−1<br />
n1<br />
n log 2.<br />
Proof: We prove it by Mathematical Induction as follows. As n 1, it holds<br />
trivially. Assume that n m holds, i.e.,<br />
.
2m<br />
s 2m ∑<br />
k1<br />
consider n m 1 as follows.<br />
x m1 <br />
2m1<br />
∑<br />
km11<br />
−1<br />
k1<br />
k<br />
1<br />
k<br />
2m<br />
∑<br />
km1<br />
1<br />
k<br />
x m<br />
x m − 1<br />
m 1 1<br />
2m 1 1<br />
2m 2<br />
s 2m 1<br />
2m 1 − 1<br />
2m 2<br />
s 2m1 .<br />
So, by Mathematical Induction, we have proved that s 2n x n for all n.<br />
By s 2n x n for all n, wehave<br />
<br />
lim n→<br />
s 2n ∑<br />
k1<br />
−1<br />
k1<br />
k<br />
log 2 lim n→<br />
x n .<br />
(c) rearrange the series in (b), writing alternately p positive terms followed by q<br />
negative terms and use (a) to show that this rearrangement has sum<br />
log 2 1 2 logp/q.<br />
∑ <br />
k1<br />
<strong>The</strong>n<br />
Proof: We prove it by using <strong>The</strong>orem 8.13. So, we can consider the new series<br />
a k as follows:<br />
a k 1<br />
2k − 1p 1 ... 1<br />
2kp − 1<br />
n<br />
S n ∑ a k<br />
k1<br />
2np<br />
∑<br />
k1<br />
np<br />
1k − ∑<br />
k1<br />
nq<br />
1<br />
2k − ∑<br />
k1<br />
1<br />
2k<br />
− 1<br />
2k − 1q ... 1<br />
2kq<br />
log 2np O 1 n − 1 2 log np − 2 O 1 n − 1 2 log nq − 2 O 1 n<br />
log 2np − log n pq O<br />
1 n<br />
log 2 p q O 1 n .<br />
So,<br />
lim n→<br />
S n log 2 1 2 logp/q<br />
by <strong>The</strong>orem 8.13.<br />
Remark: <strong>The</strong>re is a reference around rearrangement of series. <strong>The</strong> reader can see the<br />
book, Infinite Series by Chao Wen-Min, pp 216-220. (Chinese Version)<br />
(d) Find the sum of ∑ <br />
n1<br />
−1 n1 1/3n − 2 − 1/3n − 1.<br />
Proof: Write
n<br />
S n ∑−1 k1 1<br />
3k − 2 − 1<br />
3k − 1<br />
k1<br />
n<br />
∑−1 k 1<br />
k1<br />
n<br />
−∑<br />
k1<br />
−<br />
−<br />
−<br />
3n<br />
∑<br />
k1<br />
n<br />
3k − 1 ∑<br />
k1<br />
−1 3k−1 1<br />
−1 k1 1<br />
3k − 2<br />
n<br />
3k − 1 − ∑−1 3k−2 1<br />
3k − 2<br />
k1<br />
n<br />
n<br />
∑−1 3k−1 1<br />
3k − 1 ∑−1 3k−2 1<br />
3k − 2<br />
k1<br />
k1<br />
3n<br />
∑<br />
k1<br />
3n<br />
∑<br />
k1<br />
−1 k<br />
k<br />
−1 k<br />
k<br />
−1 k1<br />
k<br />
n<br />
− ∑<br />
k1<br />
− 1 3 ∑ k1<br />
n<br />
−1 3k<br />
3k<br />
n<br />
−1 k<br />
k<br />
−1 k1<br />
k<br />
− 1 3 ∑ k1<br />
→ 2 log 2.<br />
3<br />
So, the series has the sum 2 log 2.<br />
3<br />
Remark: <strong>The</strong>re is a refernece around rearrangement of series. <strong>The</strong> reader can see the<br />
book, An Introduction to Mathematical Analysis by Loo-Keng Hua, pp 323-325.<br />
(Chinese Version)<br />
8.19 Let c n a n ib n ,wherea n −1 n / n , b n 1/n 2 . Show that ∑ c n is<br />
conditioinally convergent.<br />
Proof: It is clear that ∑ c n converges. Consider<br />
∑|c n| ∑ 1 n 2 1 n 4 ∑ 1 n 1 1 n 2 ≥ ∑ 1 n<br />
Hence, ∑|c n| diverges. That is, ∑ c n is conditioinally convergent.<br />
Remark: Wesay∑ c n converges if, and only if, the real part ∑ a n converges and the<br />
imaginary part ∑ b n converges, where c n a n ib n .<br />
8.20 Use <strong>The</strong>orem 8.23 to derive the following formulas:<br />
n<br />
(a) ∑ k1<br />
logk<br />
1 k 2 log2 n A O logn<br />
n<br />
(A is constant)<br />
Proof: Letfx logx<br />
x define on 3, , then f ′ x 1−logx 0on3, . So, it is<br />
x 2<br />
clear that fx is a positive and continuous function on 3, , with<br />
log x<br />
lim x→<br />
fx lim x→ x<br />
So, by <strong>The</strong>orem 8.23, wehave<br />
lim x→<br />
1 x 0byL-Hospital Rule.
n<br />
∑<br />
k3<br />
log k<br />
k<br />
which implies that<br />
<br />
3 n log x<br />
x dx C O log n<br />
n<br />
1 2 log2 n − 1 2 log2 3 C O<br />
n<br />
∑<br />
k1<br />
log k<br />
k<br />
,whereC is a constant<br />
log n<br />
n<br />
1 2 log2 n A O<br />
,whereC is a constant<br />
log n<br />
n ,<br />
where A C log2 − 1 2 2 log2 3 is a constant.<br />
n 1<br />
1<br />
(b) ∑ k2<br />
loglog n B O (B is constant)<br />
klogk n logn<br />
1<br />
Proof: Letfx defined on 2, , then 2<br />
xlogx f′ 1<br />
x −<br />
xlogx 1 log x 0on<br />
2, . So, it is clear that fx is a positive and continuous function on 3, , with<br />
lim x→<br />
fx lim 1<br />
x→ x log x 0.<br />
So, by <strong>The</strong>orem 8.23, wehave<br />
n<br />
∑<br />
k2<br />
1<br />
k log k n<br />
2<br />
dx<br />
x log x C O 1<br />
n log n<br />
log log n B O 1<br />
n log n<br />
where B C − log log 2 is a constant.<br />
8.21 If 0 a ≤ 1, s 1, define s, a ∑ n0<br />
n a −s .<br />
,whereC is a constant<br />
,whereC is a constant<br />
(a) Show that this series converges absolutely for s 1 and prove that<br />
k<br />
∑ s, h k<br />
k<br />
s s if k 1, 2, . . .<br />
h1<br />
where s s,1 is the Riemann zeta function.<br />
Proof: First, it is clear that s, a converges absolutely for s 1. Consider<br />
k<br />
∑ s, h k<br />
h1<br />
k<br />
∑<br />
h1<br />
k<br />
∑<br />
h1<br />
<br />
∑<br />
n0<br />
<br />
∑<br />
n0<br />
k<br />
∑ ∑<br />
n0 h1<br />
<br />
k s ∑<br />
n0<br />
<br />
k s ∑<br />
n0<br />
k s s.<br />
k<br />
∑<br />
h1<br />
1<br />
n h k s<br />
k s<br />
kn h s<br />
k s<br />
kn h s<br />
1<br />
kn h s<br />
1<br />
n 1 s
(b) Prove that ∑ n1<br />
−1 n−1 /n s 1 − 2 1−s s if s 1.<br />
Proof: Let<br />
n<br />
S n ∑ j1<br />
−1 j−1<br />
j s<br />
, and thus consider its subsequence S 2n as follows:<br />
2n<br />
S 2n ∑<br />
j1<br />
2n<br />
∑<br />
j1<br />
1<br />
j s<br />
1<br />
j s<br />
n<br />
− 2 ∑<br />
j1<br />
n<br />
− 2 1−s ∑<br />
j1<br />
1<br />
2j s<br />
which implies that<br />
lim n→<br />
S 2n 1 − 2 1−s s.<br />
Since S n converges, we know that S 2n also converges and has the same value. Hence,<br />
<br />
∑<br />
n1<br />
1<br />
j s<br />
−1 n−1 /n s 1 − 2 1−s s.<br />
8.22 Given a convergent series ∑ a n , where each a n ≥ 0. Prove that ∑ a n n −p<br />
converges if p 1/2. Give a counterexample for p 1/2.<br />
Proof: Since<br />
a n n −2p<br />
2<br />
≥ a n n −2p a n n −p ,<br />
we have ∑ a n n −p converges if p 1/2 since<br />
∑ a n converges and ∑ n −2p converges if p 1/2.<br />
and<br />
For p 1/2, we consider a n <br />
1<br />
nlogn 2 , then<br />
∑ a n converges by Cauchy Condensation <strong>The</strong>orem<br />
∑ a n n −1/2 ∑ 1<br />
n log n<br />
diverges by Cauchy Condensation <strong>The</strong>orem.<br />
8.23 Given that ∑ a n diverges. Prove that ∑ na n also diverges.<br />
n<br />
Proof: Assume ∑ na n converges, then its partial sum ∑ k1<br />
ka k is bounded. <strong>The</strong>n by<br />
Dirichlet Test, we would obtain<br />
∑ka k 1<br />
k<br />
∑ a k converges<br />
which contradicts to ∑ a n diverges. Hence, ∑ na n diverges.<br />
8.24 Given that ∑ a n converges, where each a n 0. Prove that<br />
∑a n a n1 1/2<br />
also converges. Show that the converse is also true if a n is monotonic.<br />
Proof: Since<br />
a n a n1<br />
≥ a<br />
2<br />
n a n1 1/2 ,<br />
we know that<br />
∑a n a n1 1/2
converges by ∑ a n converges.<br />
Conversly, since a n is monotonic, it must be decreasing since ∑ a n converges. So,<br />
a n ≥ a n1 for all n. Hence,<br />
a n a n1 1/2 ≥ a n1 for all n.<br />
So, ∑ a n converges since ∑a n a n1 1/2 converges.<br />
8.25 Given that ∑ a n converges absolutely. Show that each of the following series<br />
also converges absolutely:<br />
(a) ∑ a n<br />
2<br />
Proof: Since∑ a n converges, then a n → 0asn → . So, given 1, there exists a<br />
positive integer N such that as n ≥ N, wehave<br />
|a n| 1<br />
which implies that<br />
a n2 |a n| for n ≥ N.<br />
So, ∑ a n2 converges if ∑|a n| converges. Of course, ∑ a n2 converges absolutely.<br />
(b) ∑ an<br />
1a n<br />
(if no a n −1)<br />
Proof: Since∑|a n| converges, we have lim n→ a n 0. So, there exists a positive<br />
integer N such that as n ≥ N, wehave<br />
1/2 |1 a n|.<br />
Hence, as n ≥ N,<br />
a n<br />
2|a<br />
1 a n|<br />
n<br />
which implies that ∑<br />
(c) ∑ a n 2<br />
1a n<br />
2<br />
Proof: It is clear that<br />
an<br />
1a n<br />
By (a), we have proved that ∑ a n 2<br />
converges. So, ∑ an<br />
1a n<br />
1a n<br />
2<br />
a2<br />
n<br />
≤ a<br />
1 a2 n2 .<br />
n<br />
converges absolutely.<br />
converges absolutely.<br />
8.26 Determine all real values of x for which the following series converges.<br />
<br />
∑<br />
n1<br />
Proof: Consider its partial sum<br />
n<br />
∑<br />
k1<br />
1 1 2 ... 1 n<br />
1 1 2<br />
... 1 k <br />
k<br />
sin nx<br />
n .<br />
sin kx<br />
as follows.<br />
As x 2m, the series converges to zero. So it remains to consider x ≠ 2m as<br />
follows. Define<br />
and<br />
a k 1 1 2<br />
... 1 k<br />
k
then<br />
b k sin kx,<br />
a k1 − a k 1 1 2 ... 1 k 1<br />
k1<br />
k 1<br />
− 1 1 2<br />
... 1 k<br />
k<br />
k1 1 ... 1 1 − k 11 1 2 k k1 2<br />
kk 1<br />
k<br />
k1<br />
<br />
− 1 1 ... 1 <br />
2 k<br />
0<br />
kk 1<br />
and<br />
n<br />
∑ b k ≤ 1<br />
k1<br />
sin x .<br />
2<br />
So, by Dirichlet Test, we know that<br />
<br />
∑<br />
k1<br />
a k b k ∑<br />
k1<br />
1 1 ... 1 <br />
2 k<br />
sin kx<br />
k<br />
... 1 k <br />
converges.<br />
From above results, we have shown that the series converges for all x ∈ R.<br />
8.27. Prove that following statements:<br />
(a) ∑ a n b n converges if ∑ a n converges and if ∑b n − b n1 converges absolutely.<br />
Proof: Consider summation by parts, i.e., <strong>The</strong>orem 8.27, then<br />
n<br />
∑<br />
k1<br />
a k b k A n b n1 − ∑<br />
k1<br />
n<br />
A k b k1 − b k .<br />
Since ∑ a n converges, then |A n| ≤ M for all n. In addition, by <strong>The</strong>orem 8.10, lim n→ b n<br />
exists. So, we obtain that<br />
(1). lim n→<br />
A n b n1 exists<br />
and<br />
n<br />
(2). ∑<br />
k1<br />
n<br />
|A k b k1 − b k | ≤ M ∑<br />
k1<br />
<br />
|b k1 − b k | ≤ M ∑<br />
k1<br />
|b k1 − b k |.<br />
(2) implies that<br />
n<br />
(3). ∑ A k b k1 − b k converges.<br />
k1<br />
n<br />
By (1) and (3), we have shown that ∑ k1<br />
a k b k converges.<br />
Remark: In 1871, Paul du Bois Reymond (1831-1889) gave the result.<br />
(b) ∑ a n b n converges if ∑ a n has bounded partial sums and if ∑b n − b n1 converges<br />
absolutely, provided that b n → 0asn → .<br />
Proof: Bysummation by parts, wehave<br />
n<br />
∑<br />
k1<br />
a k b k A n b n1 − ∑ A k b k1 − b k .<br />
k1<br />
Since b n → 0asn → and ∑ a n has bounded partial sums, say |A n| ≤ M for all n. <strong>The</strong>n<br />
n
In addition,<br />
n<br />
(2). ∑<br />
k1<br />
(1). lim n→<br />
A n b n1 exists.<br />
n<br />
|A k b k1 − b k | ≤ M ∑<br />
k1<br />
<br />
|b k1 − b k | ≤ M ∑<br />
k1<br />
|b k1 − b k |.<br />
(2) implies that<br />
n<br />
(3). ∑ A k b k1 − b k converges.<br />
k1<br />
n<br />
By (1) and (3), we have shown that ∑ k1<br />
a k b k converges.<br />
Remark: (1) <strong>The</strong> result is first discovered by Richard Dedekind (1831-1916).<br />
(2) <strong>The</strong>re is an exercise by (b), we write it as a reference. Show the convergence of the<br />
<br />
k<br />
−1<br />
series ∑ k1<br />
.<br />
k<br />
Proof: Leta k −1 k<br />
, then in order to show the convergence of<br />
1/3<br />
<br />
∑ k1<br />
−1<br />
k<br />
and b<br />
k 2/3 k 1<br />
k<br />
, it suffices to show that ∑<br />
k k1<br />
a k : S n<br />
n ∈ N, there exists j ∈ N such that j 2 ≤ N j 1 2 . Consider<br />
S n a 1 a 2 a 3 a 4 ...a 8 a 9 ....a 15 ...a j<br />
2 ...a n<br />
n<br />
is bounded sequence. Given<br />
≤ 3a 3 5a 4 7a 15 9a 16 ..4k − 1a 2k 2 −1 4k 1a 2<br />
2k<br />
if j 2k, k ≥ 2<br />
3a 3 5a 4 7a 15 9a 16 ..4k − 3a 2k−2<br />
2 if j 2k − 1, k ≥ 3<br />
then as n large enough,<br />
S n ≤ −3a 4 5a 4 −7a 16 9a 16 ... −4k − 1a 2k<br />
2 4k 1a 2k<br />
2<br />
−3a 4 5a 4 −7a 16 9a 16 ... −4k − 5a 2k−2<br />
2 4k − 3a 2k−2<br />
2<br />
which implies that as n large enough,<br />
<br />
<br />
S n ≤ 2 ∑ a 2j<br />
2 2 ∑<br />
1<br />
j2<br />
j2 2j 4/3 : M 1 *<br />
Similarly, we have<br />
M 2 ≤ S n for all n **<br />
By (*) and (**), we have shown that<br />
n<br />
a k : S n is bounded sequence.<br />
∑ k1<br />
Note: (1) By above method, it is easy to show that<br />
<br />
−1<br />
k<br />
∑<br />
k1<br />
converges for p 1/2. For 0 p ≤ 1/2, the series diverges by<br />
1<br />
n 2 p ... 1<br />
n 2 2n p ≥ 2n 1<br />
n 2 n p ≥ 2n 1<br />
n 2 n p ≥ 2n 1<br />
n 1 2p ≥ 2n n 1 1 1.<br />
k p<br />
log k<br />
−1<br />
.<br />
k<br />
<br />
(2) <strong>The</strong>re is a similar question, show the divergence of the series ∑ k1<br />
Proof: Weuse<strong>The</strong>orem 8.13 to show it by inserting parentheses as follows. We insert<br />
k<br />
−1log<br />
parentheses such that the series ∑ forms ∑−1 k b k . If we can show ∑−1 k b k<br />
k
diverges, then ∑<br />
where<br />
By (2) and (4),<br />
By (1) and (3),<br />
−1log<br />
k<br />
k<br />
also diverges. Consider<br />
b k 1 m ... 1 m r , *<br />
(1). log m N<br />
(2). logm − 1 N − 1 log em − 1 N<br />
(3). logm r N<br />
(4). logm r 1 N 1 log mr1<br />
e N<br />
m r 1<br />
e<br />
m − 1 r 1 ≥ m if m is large enough.<br />
2m ≥ r.<br />
So, as k large enough ( m is large enough),<br />
b k ≥ m r 1 r ≥ 3m m 1 by (*).<br />
3<br />
It implies that ∑−1 k b k diverges since b k does NOT tends to zero as k goes infinity.So,<br />
k<br />
−1log<br />
we have proved that the series ∑ diverges.<br />
k<br />
(3) <strong>The</strong>re is a good exercise by summation by parts, we write it as a reference.<br />
Assume that ∑ <br />
k1<br />
a k b k converges and b n ↗ with lim n→ b n . Show that b n ∑ <br />
kn<br />
a k<br />
converges.<br />
Proof: First, we show that the convergence of ∑ <br />
k1<br />
a k by Dirichlet Test as follows.<br />
Since b n ↗ , there exists a positive integer n 0 such that as n n 0 ,wehaveb n 0. So,<br />
<br />
1<br />
we have<br />
b nn0<br />
is decreasing to zero. So<br />
n1<br />
<br />
∑ a kn0<br />
k1<br />
converges by Dirichlet Test.<br />
<br />
<br />
∑a kn0 b kn0 1<br />
b kn0<br />
k1<br />
For the convergence of b n ∑ kn<br />
a k ,letn n 0 , then<br />
<br />
b n ∑ a k ∑ a k b<br />
b n k<br />
b k<br />
kn kn<br />
and define c k a k b k and d k bn<br />
b k<br />
. Note that d k is decreasing to zero. Define<br />
k<br />
C k ∑ j1<br />
c j and thus we have<br />
m m<br />
b n ∑ a k ∑ a k b<br />
b n k<br />
b k<br />
kn kn<br />
So,<br />
m<br />
∑C k − C k−1 d k<br />
kn<br />
m−1<br />
∑ C k d k − d k1 C m d m − C n−1 d n .<br />
kn<br />
.
n ∑<br />
kn<br />
<br />
a k ∑<br />
kn<br />
<br />
a k b k<br />
b n<br />
b k<br />
∑ C k d k − d k1 C d − C n−1 d n<br />
kn<br />
<br />
∑<br />
kn<br />
C k d k − d k1 − C n−1 d n<br />
by C lim k→ C k and lim k→ d k 0. In order to show the existence of lim n→ b n ∑ kn<br />
a k ,<br />
it suffices to show the existence of lim n→ ∑ <br />
kn<br />
C k d k − d k1 . Since the series<br />
∑ <br />
kn<br />
C k d k − d k1 exists, lim n→ ∑ <br />
kn<br />
C k d k − d k1 0. From above results, we have<br />
proved the convergence of lim n→ b n ∑ kn<br />
a k .<br />
Note: We also show that lim n→ b n ∑ <br />
kn<br />
a k 0 by preceding sayings.<br />
Supplement on the convergence of series.<br />
A Show the divergence of ∑ 1/k. We will give some methods listed below. If the<br />
proof is easy, we will omit the details.<br />
(1) Use Cauchy Criterion for series. Since it is easy, we omit the proof.<br />
(2) Just consider<br />
1 1 2 1 3 1 4 ... 2 1 n ≥ 1 1 2 2 1 4 1 ...2n−1 2 n<br />
1 n 2 → .<br />
Remark: We can consider<br />
1 1 ... 1 1 ... 2 10<br />
1 ...≥ 1 11 100<br />
10 9 100 90 ...<br />
Note: <strong>The</strong> proof comes from Jing Yu.<br />
(3) Use Mathematical Induction to show that<br />
1<br />
k − 1 1 k 1<br />
k 1 ≥ 3 if k ≥ 3.<br />
k<br />
<strong>The</strong>n<br />
1 1 2 1 3 1 4 1 5 1 6 ....≥ 1 3 3 3 6 3 9 ...<br />
Remark: <strong>The</strong> proof comes from Bernoulli.<br />
(4) Use Integral Test. Since the proof is easy, we omit it.<br />
(5) Use Cauchy condensation <strong>The</strong>orem. Since the proof is easy, we omit it.<br />
(6) Euler Summation Formula, the reader can give it a try. We omit the proof.<br />
(7) <strong>The</strong> reader can see the book, Princilpes of Mathematical Analysis by Walter<br />
Rudin, Exercise 11-(b) pp 79.<br />
Suppose a n 0, S n a 1 ...a n ,and∑ a n diverges.<br />
(a) Prove that ∑ an<br />
1a n<br />
diverges.<br />
Proof: Ifa n → 0asn → , then by Limit Comparison <strong>The</strong>orem, we know that<br />
diverges. If a n does not tend to zero. Claim that does not tend to zero.<br />
∑ an<br />
1a n<br />
a n<br />
1a n
Suppose NOT, it means that lim n→<br />
a n<br />
1a n<br />
0. That is,<br />
lim 1<br />
n→<br />
0 lim<br />
1 a 1 n→<br />
a n 0<br />
n<br />
which contradicts our assumption. So, ∑ an<br />
1a n<br />
(b) Prove that<br />
and deduce that ∑ an<br />
S n<br />
Proof: Consider<br />
diverges.<br />
a N1<br />
S N1<br />
a N1<br />
S N1<br />
... a Nk<br />
S Nk<br />
diverges by claim.<br />
... a Nk<br />
S Nk<br />
≥ 1 − S N<br />
S Nk<br />
then ∑ an<br />
S n<br />
diverges by Cauchy Criterion with (*).<br />
Remark: Leta n 1, then ∑ an<br />
S n<br />
(c) Prove that<br />
and deduce that ∑ an<br />
S2<br />
n<br />
Proof: Consider<br />
and<br />
So, ∑ an<br />
S n<br />
2<br />
converges.<br />
≥ a N1 ...a Nk<br />
S Nk<br />
1 − S N<br />
S Nk<br />
, *<br />
∑ 1/n diverges.<br />
a n<br />
S n<br />
2<br />
≤ 1<br />
S n−1<br />
1<br />
S n−1<br />
− 1 S n<br />
<br />
− 1 S n<br />
a n<br />
S n−1 S n<br />
≥ a n<br />
S n<br />
2 ,<br />
∑<br />
S 1<br />
n−1<br />
−<br />
S 1 n<br />
converges by telescoping series with<br />
S 1 n<br />
→ 0.<br />
converges.<br />
(d) What can be said about<br />
∑<br />
Proof: For∑<br />
the series ∑<br />
For ∑<br />
so ∑<br />
an<br />
1n 2 a n<br />
a n<br />
1 na n<br />
and ∑ a n<br />
1 n 2 a n<br />
<br />
an<br />
1na n<br />
:asa n 1 for all n, theseries∑ an<br />
1na n<br />
∑ 1<br />
1n<br />
an<br />
1na n<br />
∑ 1<br />
an<br />
1n 2 a n<br />
1k 2<br />
: Consider<br />
converges.<br />
(8) Consider ∑ sin 1 n diverges.<br />
Proof: Since<br />
a n <br />
converges.<br />
a n<br />
0ifn ≠ k2<br />
,<br />
1ifn k 2<br />
1 ≤ 1 1 n 2 a<br />
1 n a n<br />
n 2 n , 2<br />
lim<br />
sin 1 n<br />
n→ 1<br />
n<br />
1,<br />
the series ∑ 1 n diverges by Limit Comparison <strong>The</strong>orem.<br />
diverges. As
Remark: In order to show the series ∑ sin 1 n diverges, we consider Cauchy Criterion<br />
as follows.<br />
n sin 1 ≤ sin 1 ... sin 1<br />
2n n 1<br />
n n<br />
and given x ∈ R, forn 0,1,2,..., we have<br />
|sin nx| ≤ n|sin x|.<br />
So,<br />
sin 1 2 ≤ sin 1 ... sin 1<br />
n 1<br />
n n<br />
for all n. Hence, ∑ sin 1 n diverges.<br />
Note: <strong>The</strong>re are many methods to show the divergence of the series ∑ sin 1 n . We can<br />
use Cauchy Condensation <strong>The</strong>orem to prove it. Besides, by (11), it also owrks.<br />
(9) O-Stolz’s <strong>The</strong>orem.<br />
n 1<br />
Proof: LetS n ∑ j1 j<br />
and X n log n. <strong>The</strong>n by O-Stolz’s <strong>The</strong>orem, it is easy to see<br />
lim n→<br />
S n .<br />
(10) Since n k1 1 1 k<br />
diverges, the series ∑ 1/k diverges by <strong>The</strong>orem 8.52.<br />
(11) Lemma: If a n is a decreasing sequence and ∑ a n converges. <strong>The</strong>n<br />
lim n→ na n 0.<br />
Proof: Sincea n → 0anda n is a decreasing sequence, we conclude that a n ≥ 0.<br />
Since ∑ a n converges, given 0, there exists a positive integer N such that as n ≥ N,<br />
we have<br />
a n ..a nk /2 for all k ∈ N<br />
which implies that<br />
k 1a nk /2 since a n ↘.<br />
Let k n, then as n ≥ N, wehave<br />
n 1a 2n /2<br />
which implies that as n ≥ N<br />
2n 1a 2n <br />
which implies that<br />
lim n→<br />
2na 2n 0 since lim n→<br />
a n 0. *<br />
Similarly, we can show that<br />
lim n→<br />
2n 1a 2n1 0. **<br />
So,by(*)adn(**),wehaveprovedthatlim n→ na n 0.<br />
Remark: From this, it is clear that ∑ 1 n diverges. In addition, we have the convergence<br />
of ∑ na n − a n1 . We give it a proof as follows.<br />
Proof: Write
then<br />
since<br />
n<br />
lim n→<br />
∑<br />
k1<br />
n<br />
S n ∑ ka k − a k1 <br />
k1<br />
n<br />
∑<br />
k1<br />
a k − na n1 ,<br />
lim n→<br />
S n exists<br />
a k exists and lim n→<br />
na n 0.<br />
B Prove that ∑ 1 p diverges, where p is a prime.<br />
Proof: GivenN, letp 1 ,...,p k be the primes that divide at least one integer≤ N. <strong>The</strong>n<br />
N<br />
∑<br />
n1<br />
k<br />
1<br />
n ≤ <br />
j1<br />
k<br />
<br />
j1<br />
≤ exp<br />
1 1 p j<br />
1 p j<br />
2 ...<br />
1<br />
1 − p 1 j<br />
by 1 − x −1 ≤ e 2x if 0 ≤ x ≤ 1/2. Hence, ∑ 1 p diverges since ∑ 1 n diverges.<br />
Remark: <strong>The</strong>re are many proofs about it. <strong>The</strong> reader can see the book, An<br />
Introduction To <strong>The</strong> <strong>The</strong>ory Of <strong>Number</strong>s by Loo-Keng Hua, pp 91-93. (Chinese<br />
Version)<br />
C Discuss some series related with ∑ sink .<br />
k<br />
STUDY: (1) We have shown that the series ∑ sin 1 diverges.<br />
k<br />
(2) <strong>The</strong> series ∑ sinna b diverges where a ≠ n for all n ∈ Z and b ∈ R.<br />
Proof: Suppose that ∑ sinna b converges, then lim n→ sinna b 0. Hence,<br />
lim n→|sinn 1a b − sinna b| 0. Consider<br />
|sinn 1a b − sinna b|<br />
k<br />
∑<br />
j1<br />
2<br />
p j<br />
2cos na b a 2<br />
sin a 2<br />
which implies that<br />
2 cosna b cos a<br />
2<br />
− sinna b sin a 2<br />
sin a 2
lim n→<br />
|sinn 1a b − sinna b|<br />
lim n→<br />
sinn 1a b − sinna b<br />
lim n→<br />
sup 2 cosna b cos a<br />
2<br />
− sinna b sin a 2<br />
sin a 2<br />
lim n→<br />
sup 2 cosna b cos a sin a 2 2<br />
|sin a| ≠ 0<br />
which is impossible. So, ∑ sinna b diverges.<br />
Remark: (1) By the same method, we can show the divergence of ∑ cosna b if<br />
a ≠ n for all n ∈ Z and b ∈ R.<br />
(2) <strong>The</strong> reader may give it a try to show that,<br />
and<br />
p<br />
by considering ∑ n0<br />
(**).<br />
(3) <strong>The</strong> series ∑ sink<br />
k<br />
p<br />
∑<br />
n0<br />
cosna b <br />
Proof: First, it is clear that ∑ sink<br />
k<br />
|∑ sin k| ≤<br />
1<br />
sin 1 2<br />
partial sums as follows: Since<br />
sin<br />
p1<br />
2 b<br />
sin b 2<br />
sin a p 2 b *<br />
p<br />
p1<br />
sin<br />
∑ sinna b b 2<br />
cos a p sin b 2 b **<br />
n0<br />
2<br />
e inab . However, it is not easy to show the divergence by (*) and<br />
converges conditionally.<br />
converges by Dirichlet’s Test since<br />
, we consider its<br />
. In order to show that the divergence of ∑ sink<br />
k<br />
3n3<br />
n<br />
∑ sin k ∑<br />
sin 3k 1 sin 3k 2 sin 3k 3<br />
k<br />
3k 1 3k 2 3k 3<br />
k1<br />
k0<br />
and note that there is one value is bigger than 1/2 among three values |sin 3k 1|,<br />
|sin 3k 2|, and|sin 3k 3|. So,<br />
3n3<br />
∑<br />
k1<br />
sin k<br />
k<br />
n 1<br />
2<br />
≥ ∑<br />
3k 3<br />
k0<br />
which implies the divergence of ∑ sink .<br />
k<br />
<br />
Remark: <strong>The</strong> series is like Dirichlet Integral sinx<br />
0<br />
x<br />
Integral converges conditionally.<br />
(4) <strong>The</strong> series ∑ |sink|r diverges for any r ∈ R.<br />
k<br />
Proof: We prove it by three cases as follows.<br />
(a) As r ≤ 0, we have<br />
So, ∑ |sink|r diverges in this case.<br />
k<br />
(b) As 0 r ≤ 1, we have<br />
∑<br />
|sin k|r<br />
k<br />
≥ ∑ 1 k .<br />
dx. Also, we know that Dirichlet
|sin k|r<br />
∑<br />
k<br />
So, ∑ |sink|r diverges in this case by (3).<br />
k<br />
(c) As r 1, we have<br />
3n3<br />
∑ |sin k|<br />
r<br />
k<br />
k1<br />
n<br />
∑<br />
k0<br />
n<br />
≥ ∑<br />
k0<br />
|sin 3k 1| r<br />
3k 1<br />
1 2 r<br />
3k 3 .<br />
≥ ∑<br />
<br />
|sin k|<br />
k<br />
.<br />
|sin 3k 2|r<br />
3k 2<br />
<br />
|sin 3k 3|r<br />
3k 3<br />
So, ∑ |sink|r diverges in this case.<br />
k<br />
(5) <strong>The</strong> series ∑ sin2p−1 k<br />
,wherep ∈ N, converges.<br />
k<br />
Proof: We will prove that there is a positive integer Mp such that<br />
n<br />
∑<br />
k1<br />
sin 2p−1 k ≤ Mp for all n. *<br />
So, if we can show (*), then by Dirichlet’s Test, we have proved it. In order to show (*),<br />
we claim that sin 2p−1 k can be written as a linear combination of sink, sin3k,...,<br />
sin2p − 1k. So,<br />
n<br />
∑<br />
k1<br />
n<br />
sin 2p−1 k ∑<br />
k1<br />
n<br />
≤ |a 1 | ∑<br />
k1<br />
|a 1 |<br />
a 1 sin k a 2 sin 3k ...a p sin2p − 1k<br />
n<br />
sin k ...|a p| ∑ sin2p − 1k<br />
k1<br />
|a<br />
≤ ... p|<br />
: Mp by <strong>The</strong>orem 8.30.<br />
sin 1 2 sin 2p−1<br />
2<br />
We show the claim by Mathematical Induction as follows. As p 1, it trivially holds.<br />
Assume that as p s holds, i.e.,<br />
then as p s 1, we have<br />
sin 2s−1 k ∑<br />
j1<br />
s<br />
a j sin2j − 1k
sin 2s1 k sin 2 ksin k 2s−1<br />
sin 2 k<br />
s<br />
∑ a j sin2j − 1k<br />
j1<br />
s<br />
∑ a j sin 2 k sin2j − 1k<br />
j1<br />
s<br />
∑ a j<br />
j1<br />
1 2<br />
1 2<br />
s<br />
∑<br />
j1<br />
1 − cos2k<br />
2<br />
sin2j − 1k<br />
by induction hypothesis<br />
s<br />
a j sin2j − 1k − ∑ a j cos2k sin2j − 1k<br />
j1<br />
s<br />
∑ a j sin2j − 1k − 1 2 ∑ s<br />
j1<br />
j1<br />
a j sin2j 1k sin2j − 3k<br />
which is a linear combination of sink,...,sin2s 1k. Hence, we have proved the claim<br />
by Mathematical Induction.<br />
Remark: By the same argument, the series<br />
n<br />
∑<br />
k1<br />
cos 2p−1 k<br />
is also bounded, i.e., there exists a positive number Mp such that<br />
n<br />
(6) Define ∑ k1<br />
n<br />
∑<br />
k1<br />
|cos 2p−1 k| ≤ Mp.<br />
sinkx<br />
: F<br />
k n x, then F n x is boundedly convergent on R.<br />
Proof: SinceF n x is a periodic function with period 2, andF n x is an odd function.<br />
So, it suffices to consider F n x is defined on 0, . In addition, F n 0 0 for all n.<br />
Hence, the domain I that we consider is 0, . Note that sinkx x<br />
cosktdt. So,<br />
0<br />
n<br />
F n x ∑<br />
k1<br />
<br />
0<br />
x<br />
∑<br />
k1<br />
<br />
0<br />
x<br />
<br />
0<br />
x<br />
sin kx<br />
k<br />
n<br />
cosktdt<br />
sinn 1 t − sin 1 t<br />
2 2<br />
2sin 1 t dt<br />
2<br />
sinn 1 2 t<br />
t<br />
dt <br />
0<br />
x<br />
1<br />
2sin t 2<br />
− 1 t<br />
k<br />
sin n 1 2 t dt − x 2<br />
which implies that<br />
<br />
0<br />
n 1 2<br />
x sin t<br />
t<br />
|F n x| ≤ <br />
0<br />
n 1 2 x sin t<br />
t<br />
dt <br />
0<br />
x t − 2sin t 2<br />
2t sin t 2<br />
dt <br />
0<br />
x t − 2sin t 2<br />
2t sin t 2<br />
sin n 1 2 t dt − x 2<br />
sin n 1 2 t dt 2 .
n<br />
For the part 1 2<br />
0<br />
that<br />
x sint<br />
<br />
t<br />
dt :Since sint<br />
0 t<br />
dt converges, there exists a positive M 1 such<br />
<br />
0<br />
n 1 2<br />
x sin t<br />
t<br />
dt ≤ M 1 for all x ∈ I and for all n.<br />
For the part 0<br />
x t−2sin t 2<br />
2t sin t 2<br />
sinn 1 2 tdt<br />
: Consider<br />
<br />
0<br />
x t − 2sin t 2<br />
2t sin t 2<br />
sin n 1 2<br />
t dt<br />
≤ <br />
0<br />
x t − 2sin t 2<br />
2t sin t 2<br />
dt since t − 2sin t 2 0onI<br />
≤ <br />
0<br />
t − 2sin t 2<br />
2t sin t 2<br />
dt : M 2 since lim<br />
t→0<br />
<br />
t − 2sin t 2<br />
2t sin t 2<br />
Hence,<br />
|F n x| ≤ M 1 M 2 for all x ∈ I and for all n.<br />
2<br />
So, F n x is uniformly bounded on I. It means that F n x is uniformly bounded on R.<br />
In addition, since<br />
F n x <br />
0<br />
n 1 2<br />
fixed x ∈ I, wehave<br />
x sin t<br />
t<br />
dt <br />
0<br />
x t − 2sin t 2<br />
2t sin t 2<br />
sin t<br />
0 t<br />
dt exists.<br />
and by Riemann-Lebesgue Lemma, in the text book, pp 313,<br />
0.<br />
sin n 1 2 t dt − x 2 ,<br />
x t − 2sin<br />
lim n→<br />
t 2<br />
0 2t sin t sin n 1 t dt 0.<br />
2<br />
2<br />
So, we have proved that<br />
lim n→<br />
F n x sin t<br />
0 t<br />
dt −<br />
2 x where x ∈ 0, .<br />
Hence, F n x is pointwise convergent on I. It means that F n x is pointwise<br />
convergent on R.<br />
Remark: (1) For definition of being boundedly convergent on a set S, the reader can<br />
see the text book, pp 227.<br />
(2) In the proof, we also shown the value of Dirichlet Integral<br />
sin t<br />
0 t<br />
dt 2<br />
by letting x .<br />
(3) <strong>The</strong>re is another proof on uniform bound. We write it as a reference.<br />
Proof: <strong>The</strong> domain that we consider is still 0, . Let 0, and consider two cases as<br />
follows.<br />
(a) x ≥ 0 : Using summation by parts,
n<br />
∑<br />
k1<br />
sin kx<br />
k<br />
≤ 1<br />
n 1 ∑ n<br />
k1<br />
≤ 1<br />
n 1<br />
sin kx<br />
k<br />
n<br />
∑<br />
k1<br />
1<br />
sin 2 1<br />
sin 2 1 − 1<br />
n 1<br />
1<br />
sin .<br />
2<br />
(b) 0 x ≤ :LetN 1 x , consider two cases as follows.<br />
As n N, then<br />
and as n ≥ N, then<br />
≤<br />
n<br />
∑<br />
k1<br />
N−1<br />
∑<br />
k1<br />
sin kx<br />
k<br />
sin kx<br />
k<br />
n<br />
≤ 1 ∑<br />
kN<br />
sin kx<br />
k<br />
≤ 1 1<br />
n 1 ∑ n<br />
k1<br />
n<br />
∑<br />
kN<br />
by (*)<br />
sin kx<br />
k<br />
by summation by parts<br />
≤ 1 1<br />
n 1 sin x 2<br />
1 2 .<br />
1 x sin x 2<br />
2<br />
Note that lim x→0<br />
<br />
1 x sin x 2<br />
n<br />
∑<br />
k1<br />
sin kx<br />
k<br />
<br />
1<br />
N sin x 2<br />
sin kx<br />
k<br />
N−1<br />
1<br />
N ∑ k1<br />
k<br />
∑ sin jx 1<br />
k 1 − 1 k<br />
j1<br />
≤ n|x| N|x| ≤ 1 *<br />
sin kx<br />
k<br />
1<br />
N<br />
− 1<br />
n 1<br />
n<br />
∑<br />
kN<br />
1<br />
sin x 2<br />
4. So, we may choose a ′ such that<br />
k<br />
∑ sin jx 1<br />
k 1 − 1 k<br />
j1<br />
2 ≤ 5 for all x ∈ 0, <br />
1 x sin x ′ .<br />
2<br />
By preceding sayings, we have proved that F n x is uniformly bounded on I. It means<br />
that F n x is uniformly bounded on R.<br />
D In 1911, Otto Toeplitz proves the following. Let a n and x n be two sequences<br />
1<br />
such that a n 0 for all n with lim n→ a 1 ...a n<br />
0 and lim n→ x n x. <strong>The</strong>n<br />
lim<br />
a 1 x 1 ...a n x n<br />
n→ a 1 ...a n<br />
x.<br />
n<br />
Proof: LetS n ∑ k1<br />
n<br />
a k and T n ∑ k1<br />
a k x k , then<br />
lim<br />
T n1 − T n<br />
n→ S n1 − S n<br />
lim n→<br />
a n1 x n1<br />
a n1<br />
lim n→<br />
x n1 x.<br />
So, by O-Stolz’s <strong>The</strong>orem, wehaveproveit.<br />
Remark: (1) Let a n 1, then it is an extension of <strong>The</strong>orem 8.48.<br />
(2) Show that
lim n→<br />
sin ... sin n<br />
1 ... 1 n<br />
.<br />
Proof: Write<br />
sin ... sin n<br />
1 1sin ... 1 1 n n sin n<br />
,<br />
1 ... 1 n<br />
1 ... 1 n<br />
the by Toeplitz’s <strong>The</strong>orem, we have proved it.<br />
E <strong>The</strong>orem 8.16 emphasizes the decrease of the sequence a n , we may ask if we<br />
remove the condition of decrease, is it true <strong>The</strong> answer is NOT necessary. For example,<br />
let<br />
a n 1 n −1n1 . 0<br />
2n<br />
F Some questions on series.<br />
(1) Show the convergence of the series ∑ n1<br />
log n sin 1 n .<br />
Proof: Since n sin 1 n 1 for all n, logn sin 1 n 0 for all n. Hence, we consider the<br />
new series<br />
<br />
∑<br />
n1<br />
− log n sin 1 <br />
n ∑ log<br />
n1<br />
sin 1/n<br />
1/n<br />
as follows. Let a n log sin1/n and b<br />
1/n n log 1 1 , then<br />
n 2<br />
lim<br />
a n<br />
n→<br />
<br />
b 1 n 6 .<br />
In addition,<br />
∑ b n ≤ ∑ 1 n 2<br />
by e x ≥ 1 x for all x ∈ R. From the convergence of ∑ b n , we have proved that the<br />
convergence of ∑ a n by Limit Comparison Test.<br />
<br />
∑ n1<br />
(2) Suppose that a n ∈ R, and the series ∑ <br />
n1<br />
converges absolutely.<br />
a n<br />
n<br />
Proof: By A. P. ≥ G. P. , we have<br />
a n2 1 n 2<br />
2<br />
<br />
which implies that ∑ n1<br />
a n<br />
n<br />
≥<br />
converges absolutely.<br />
a n2 converges. Prove that the series<br />
Remark: We metion that there is another proof by using Cauchy-Schwarz inequality.<br />
the difference of two proofs is that one considers a n , and another considers the partial<br />
sums S n .<br />
Proof: ByCauchy-Schwarz inequality,<br />
<br />
which implies that ∑ n1<br />
a n<br />
n<br />
n<br />
∑<br />
k1<br />
|a n|<br />
k<br />
2<br />
≤<br />
a n<br />
n<br />
∑ a2<br />
k<br />
k1<br />
converges absolutely.<br />
Double sequences and series<br />
8.28 Investigate the existence of the two iterated limits and the double limit of the<br />
n<br />
∑<br />
k1<br />
1<br />
k 2
double sequence f defined by the followings. Answer. Double limit exists in (a), (d), (e),<br />
(g). Both iterated limits exists in (a), (b), (h). Only one iterated limit exists in (c), (e).<br />
Neither iterated limit exists in (d), (f).<br />
(a) fp, q pq<br />
1<br />
Proof: It is easy to know that the double limit exists with lim p,q→ fp, q 0by<br />
definition. We omit it. In addition, lim p→ fp, q 0. So, lim q→ lim p→ fp, q 0.<br />
Similarly, lim p→ lim q→ fp, q 0. Hence, we also have the existence of two iterated<br />
limits.<br />
(b) fp, q <br />
p<br />
pq<br />
Proof: Letq np, then fp, q 1 . It implies that the double limit does not exist.<br />
n1<br />
However, lim p→ fp, q 1, and lim q→ fp, q 0. So, lim q→ lim p→ fp, q 1, and<br />
lim p→ lim q→ fp, q 0.<br />
(c) fp, q −1p p<br />
pq<br />
Proof: Letq np, then fp, q −1p<br />
n1<br />
. It implies that the double limit does not exist.<br />
In addition, lim q→ fp, q 0. So, lim p→ lim q→ fp, q 0. However, since<br />
lim p→ fp, q does not exist, lim q→ lim p→ fp, q does not exist.<br />
(d) fp, q −1 pq 1 p 1 q <br />
Proof: It is easy to know lim p,q→ fp, q 0. However, lim q→ fp, q and lim p→ fp, q<br />
do not exist. So, neither iterated limit exists.<br />
(e) fp, q −1p<br />
q<br />
Proof: It is easy to know lim p,q→ fp, q 0. In addition, lim q→ fp, q 0. So,<br />
lim p→ lim q→ fp, q 0. However, since lim p→ fp, q does not exist,<br />
lim q→ lim p→ fp, q does not exist.<br />
(f) fp, q −1 pq<br />
Proof: Let p nq, then fp, q −1 n1q . It means that the double limit does not<br />
exist. Also, since lim p→ fp, q and lim q→ fp, q do not exist, lim q→ lim p→ fp, q and<br />
lim p→ lim q→ fp, q do not exist.<br />
(g) fp, q cos p<br />
q<br />
Proof: Since|fp, q| ≤ 1 q , then lim p,q→ fp, q 0, and lim p→ lim q→ fp, q 0.<br />
However, since cosp : p ∈ N dense in −1, 1, we know that lim q→ lim p→ fp, q does<br />
not exist.<br />
(h) fp, q p q<br />
∑<br />
q 2 n1<br />
sin n p<br />
Proof: Rewrite<br />
and thus let p nq, fp, q sin 1 2n sin<br />
fp, q p sin q<br />
2p<br />
q1<br />
2nq<br />
nqsin 1<br />
2nq<br />
sin<br />
q1<br />
2p<br />
q 2 sin 1<br />
2p<br />
. It means that the double limit does not exist.<br />
However, lim p→ fp, q q1<br />
2q<br />
since sin x~x as x → 0. So, lim q→ lim p→ fp, q 1 2 .
Also, lim q→ fp, q lim q→ p sin 1 2p<br />
lim p→ lim q→ fp, q 0.<br />
8.29 Prove the following statements:<br />
sin q 2p<br />
sin<br />
q1<br />
2p<br />
q 2<br />
0since|sin x| ≤ 1. So,<br />
(a) A double series of positive terms converges if, and only if, the set of partial sums is<br />
bounded.<br />
Proof: ()Suppose that ∑ m,n<br />
fm, n converges, say ∑ m,n<br />
fm, n A 1 , then it means<br />
that lim p,q→ sp, q A 1 . Hence, given 1, there exists a positive integer N such that as<br />
p, q ≥ N, wehave<br />
|sp, q| ≤ |A 1 | 1.<br />
So, let A 2 maxsp, q :1≤ p, q N, wehave|sp, q| ≤ maxA 1 , A 2 for all p, q.<br />
Hence, we have proved the set of partial sums is bounded.<br />
()Suppose that the set of partial sums is bounded by M, i.e., if<br />
S sp, q : p, q ∈ N, then sup S : A ≤ M. Hence, given 0, then there exists a<br />
sp 1 , q 1 ∈ S such that<br />
A − sp 1 , q 1 ≤ A.<br />
Choose N maxp 1 , q 1 , then<br />
A − sp, q ≤ A for all p, q ≥ N<br />
since every term is positive. Hence, we have proved lim p,q→ sp, q A. Thatis,<br />
∑ m,n<br />
fm, n converges.<br />
(b) A double series converges if it converges absolutely.<br />
p<br />
Proof: Lets 1 p, q ∑ m1<br />
q<br />
∑ n1<br />
p<br />
|fm, n| and s 2 p, q ∑ m1<br />
q<br />
∑ n1<br />
fm, n, wewant<br />
to show that the existence of lim p,q→ s 2 p, q by the existence of lim p,q→ s 1 p, q as<br />
follows.<br />
Since lim p,q→ s 1 p, q exists, say its limit a. <strong>The</strong>n lim p→ s 1 p, p a. It implies that<br />
lim p→ s 2 p, p converges, say its limit b. So, given 0, there exists a positive integer N<br />
such that as p, q ≥ N<br />
|s 1 p, p − s 1 q, q| /2<br />
and<br />
|s 2 N, N − b| /2.<br />
So, as p ≥ q ≥ N,<br />
|s 2 p, q − b| |s 2 N, N − b s 2 p, q − s 2 N, N|<br />
/2 |s 2 p, q − s 2 N, N|<br />
/2 s 1 p, p − s 1 N, N<br />
/2 /2<br />
.<br />
Similarly for q ≥ p ≥ N. Hence, we have shown that<br />
lim<br />
p,q→ s 2p, q b.<br />
That is, we have prove that a double series converges if it converges absolutely.<br />
(c) ∑ m,n<br />
e −m2 n 2 <br />
converges.<br />
Proof: Letfm, n e −m2 n 2 <br />
, then by <strong>The</strong>orem 8.44, we have proved that
∑ m,n<br />
e −m2 n 2 <br />
converges since ∑ m,n<br />
e −m2 n 2 <br />
∑ m<br />
e −m2 ∑ n<br />
e −n2 .<br />
<br />
Remark: ∑ m,n1<br />
e −m2 n 2 <br />
∑ <br />
m1<br />
e −m2 ∑ n1<br />
e −n2 e<br />
<br />
e 2 −1<br />
8.30 Asume that the double series ∑ m,n<br />
anx mn converges absolutely for |x| 1. Call<br />
its sum Sx. Show that each of the following series also converges absolutely for |x| 1<br />
and has sum Sx :<br />
<br />
∑<br />
n1<br />
Proof: By<strong>The</strong>orem 8.42,<br />
So, ∑ <br />
n1<br />
∑ <br />
n1<br />
that<br />
∑<br />
m,n<br />
<br />
an x n<br />
1 − x n , ∑<br />
<br />
anx mn ∑<br />
n1<br />
n1<br />
<br />
an ∑<br />
m1<br />
Anx n ,whereAn ∑ ad.<br />
d|n<br />
<br />
x mn ∑ an<br />
n1<br />
2<br />
.<br />
x n<br />
1 − xn if |x| 1.<br />
an xn converges absolutely for<br />
1−x<br />
|x| 1 and has sum Sx.<br />
n<br />
Since every term in ∑ m,n<br />
anx mn , the term appears once and only once in<br />
Anx n . <strong>The</strong> converse also true. So, by <strong>The</strong>orem 8.42 and <strong>The</strong>orem 8.13, we know<br />
<br />
∑<br />
n1<br />
Anx n ∑ anx mn Sx.<br />
m,n<br />
8.31 If is real, show that the double series ∑ m,n<br />
m in − converges absolutely if,<br />
p q<br />
and only if, 2. Hint. Let sp, q ∑ m1<br />
∑ n1<br />
|m in| − . <strong>The</strong> set<br />
m in : m 1,2,...p, n 1, 2, . . . , p<br />
consists of p 2 complex numbers of which one has absolute value 2 , three satisfy<br />
|1 2i| ≤ |m in| ≤ 2 2, five satisfy |1 3i| ≤ |m in| ≤ 3 2, etc. Verify this<br />
geometricall and deduce the inequlity<br />
p<br />
p<br />
2 −/2 ∑<br />
2n − 1<br />
n ≤ sp, p ≤ ∑<br />
2n − 1<br />
n1<br />
n1 n 2 1 . /2<br />
Proof: Since the hint is trivial, we omit the proof of hint. From the hint, we have<br />
p<br />
∑<br />
n1<br />
2n − 1<br />
n 2 ≤ sp, p ∑<br />
m1<br />
p<br />
p<br />
∑<br />
n1<br />
p<br />
|m in| − ≤ ∑<br />
n1<br />
2n − 1<br />
1 n 2 /2 .<br />
Thus, it is clear that the double series ∑ m,n<br />
m in − converges absolutely if, and only if,<br />
2.<br />
8.32 (a) Show that the Cauchy product of ∑ <br />
n0<br />
−1 n1 / n 1 with itself is a<br />
divergent series.<br />
Proof: Since
n<br />
c n ∑ a k b n−k<br />
k0<br />
n<br />
∑<br />
k0<br />
−1 k<br />
k 1<br />
n<br />
−1 n ∑<br />
k0<br />
−1 n−k<br />
n − k 1<br />
1<br />
k 1 n − k 1<br />
and let fk n − k 1k 1 −k − n 2 2 n2<br />
2 2 ≤ n2 for k 0, 1, . . . , n.<br />
2<br />
Hence,<br />
n<br />
|c n| ∑<br />
k0<br />
1<br />
k 1 n − k 1<br />
2n 1<br />
≥ → 2asn → .<br />
n 2<br />
That is, the Cauchy product of ∑ <br />
n0<br />
−1 n1 / n 1 with itself is a divergent series.<br />
(b) Show that the Cauchy product of ∑ <br />
n0<br />
−1 n1 /n 1 with itself is the series<br />
<br />
2 ∑ −1<br />
n1<br />
1 <br />
n 1<br />
1 2 ... 1 n .<br />
n1<br />
Proof: Since<br />
n<br />
c n ∑ a k b n−k<br />
we have<br />
<br />
∑<br />
n0<br />
k0<br />
n<br />
∑<br />
k0<br />
−1 n ∑<br />
k0<br />
2−1n<br />
n 2<br />
−1 n<br />
n − k 1k 1<br />
<br />
c n ∑<br />
n<br />
n0<br />
<br />
2 ∑<br />
n0<br />
<br />
2 ∑<br />
n1<br />
∑ k0<br />
1<br />
n 2<br />
n<br />
2−1<br />
n<br />
n 2<br />
1<br />
k 1 ,<br />
∑ n<br />
k0<br />
1<br />
k 1 1<br />
n − k 1<br />
1<br />
k 1<br />
−1<br />
n<br />
n 2 1 1 2 ... 1<br />
n 1<br />
−1<br />
n1<br />
n 1<br />
(c) Does this converge Why<br />
Proof: Yes by the same argument in Exercise 8.26.<br />
1 1 2 ... 1 n .<br />
8.33 Given two absolutely convergent power series, say ∑ <br />
n0<br />
a n x n and ∑ n0<br />
b n x n ,<br />
having sums Ax and Bx, respectively, show that ∑ n0<br />
c n x n AxBx where
n<br />
c n ∑ a k b n−k .<br />
k0<br />
Proof: By<strong>The</strong>orem 8.44 and <strong>The</strong>orem 8.13, it is clear.<br />
Remark: We can use Mertens’ <strong>The</strong>orem, thenitisclear.<br />
8.34 Aseriesoftheform∑ <br />
n1<br />
a n /n s is called a Dirichlet series. Given two absolutely<br />
convergent Dirichlet series, say ∑ <br />
n1<br />
a n /n s and ∑ <br />
n1<br />
b n /n s , having sums As and Bs,<br />
respectively, show that ∑ n1<br />
c n /n s AsBs, wherec n ∑ d|n<br />
a d b n/d .<br />
Proof: By<strong>The</strong>orem 8.44 and <strong>The</strong>orem 8.13, wehave<br />
<br />
∑<br />
n1<br />
a n /n s<br />
<br />
∑<br />
n1<br />
<br />
b n /n s ∑<br />
n1<br />
where<br />
C n ∑ a d d −s b n/d n/d −s<br />
d|n<br />
n −s ∑ a d b n/d<br />
d|n<br />
c n /n s .<br />
So,wehaveprovedit.<br />
8.35 s ∑ <br />
n1<br />
1/n s , s 1, show that 2 s ∑ <br />
n1<br />
dn/n s ,wheredn is the<br />
number of positive divisors of n (including 1 and n).<br />
Proof: ItisclearbyExercise 8.34. So, we omit the proof.<br />
Ces’aro summability<br />
8.36 Show that each of the following series has C,1 sum 0 :<br />
(a) 1 − 1 − 1 1 1 − 1 − 1 1 1 −−.<br />
Proof: It is clear that |s 1 ...s n| ≤ 1 for all n, wheres n means that the nth partial sum<br />
of given series. So,<br />
s 1 ...s n<br />
n ≤ 1 n<br />
which implies that the given series has C,1 sum 0.<br />
(b) 1 − 1 1 1 − 1 1 1 − 1 − .<br />
2 2 2 2 2<br />
Proof: It is clear that |s 1 ...s n| ≤ 1 for all n, wheres 2<br />
n means that the nth partial<br />
sum of given series. So,<br />
s 1 ...s n<br />
n ≤<br />
2n<br />
1<br />
which implies that the given series has C,1 sum 0.<br />
(c) cos x cos3x cos5x (x real, x ≠ m).<br />
Proof: Lets n cosx ... cos2n − 1x, then<br />
C n
So,<br />
n<br />
s n ∑ cos2k − 1x<br />
j1<br />
sin 2nx<br />
2sinx .<br />
n<br />
∑ j1<br />
s ∑ n<br />
j sin 2jx<br />
j1<br />
n <br />
2n sin x<br />
sin nx sinn 1x<br />
<br />
2n sin x sin x<br />
≤ 1 → 0<br />
2nsin x 2<br />
which implies that the given series has C,1 sum 0.<br />
8.37 Given a series ∑ a n ,let<br />
Prove that:<br />
(a) t n n 1s n − n n<br />
n<br />
s n ∑<br />
k1<br />
Proof: DefineS 0 0, and thus<br />
n<br />
t n ∑ ka k<br />
k1<br />
n<br />
n<br />
a k , t n ∑<br />
k1<br />
∑ ks k − s k−1 <br />
k1<br />
n<br />
∑<br />
k1<br />
n<br />
n<br />
ks k − ∑ ks k−1<br />
k1<br />
∑ ks k − ∑k 1s k<br />
k1<br />
n<br />
∑<br />
k1<br />
n−1<br />
k1<br />
n<br />
ks k − ∑<br />
k1<br />
n 1s n − ∑<br />
k1<br />
n 1s n − n n .<br />
ka k , n 1 n<br />
n ∑ s k .<br />
k1<br />
k 1s k n 1s n<br />
(b) If ∑ a n is C,1 summable, then ∑ a n converges if, and only if, t n on as<br />
n → .<br />
Proof: Assume that ∑ a n converges. <strong>The</strong>n lim n→ s n exists, say its limit a. By(a),we<br />
have<br />
t nn n 1 n s n − n .<br />
<strong>The</strong>n by <strong>The</strong>orem 8.48, we also have lim n→ n a. Hence,<br />
n<br />
s k
lim<br />
t nn<br />
n→<br />
lim n 1<br />
n→ n s n − n<br />
lim n 1<br />
n→ n lim n→<br />
s n − lim n→<br />
n<br />
1 a − a<br />
0<br />
which is t n on as n → .<br />
Conversely, assume that t n on as n → , thenby(a),wehave<br />
n t nn n<br />
n 1 n 1 n s n<br />
which implies that (note that lim n→ n exists by hypothesis)<br />
lim n→<br />
s n lim n t nn<br />
n→<br />
n<br />
n 1 n 1 n<br />
lim n<br />
n→ n 1 lim t nn lim n<br />
n→ n→ n 1 lim<br />
n→<br />
n<br />
1 0 1 lim n→<br />
n<br />
lim n→<br />
n<br />
That is, ∑ a n converges.<br />
(c) ∑ a n is C,1 summable if, and only if, ∑ t n /nn 1 converges.<br />
Proof: Consider<br />
which implies that<br />
t n<br />
nn 1 s n −<br />
n<br />
∑<br />
k1<br />
n<br />
n 1<br />
n n − n − 1 n−1<br />
n − n<br />
n 1<br />
n<br />
n 1 n − n − n<br />
1 n−1<br />
t k<br />
kk 1 <br />
n<br />
n 1 n. *<br />
()Suppose that ∑ a n is C,1 summable, i.e., lim n→ n exists. <strong>The</strong>n<br />
n t<br />
lim n→ ∑ k<br />
k1<br />
exists by (*).<br />
kk1<br />
n t<br />
()Suppose that lim n→ ∑ k<br />
k1<br />
exists. <strong>The</strong>n lim<br />
kk1<br />
n→ n exists by (*). Hence, ∑ a n<br />
is C,1 summable.<br />
8.38 Given a monotonic a n of positive terms, such that lim n→ a n 0. Let<br />
n<br />
s n ∑<br />
k1<br />
Prove that:<br />
(a) v n 1 u 2 n −1 n s n /2.<br />
n<br />
a k , u n ∑<br />
k1<br />
Proof: Defines 0 0, and thus consider<br />
−1 k a k , v n ∑<br />
k1<br />
n<br />
−1 k s k .
n<br />
u n ∑−1 k a k<br />
k1<br />
n<br />
∑−1 k s k − s k−1 <br />
k1<br />
n<br />
∑−1 k s k ∑−1 k1 s k−1<br />
k1<br />
n<br />
∑<br />
k1<br />
n<br />
k1<br />
n<br />
−1 k s k ∑<br />
k1<br />
−1 k s k −1 n1 s n<br />
which implies that<br />
2v n −1 n1 s n<br />
v n 1 2 u n −1 n s n /2.<br />
(b) ∑ <br />
n1<br />
−1 n s n is C,1 summable and has Ces’aro sum 1 ∑ −1 n a<br />
2 n1 n .<br />
Proof: First, lim n→ u n exists since it is an alternating series. In addition, since<br />
lim n→ a n 0, we know that lim n→ s n /n 0by<strong>The</strong>orem 8.48. Hence,<br />
Consider by (a),<br />
v n<br />
n<br />
∑ k1<br />
v k<br />
n<br />
u n<br />
2n −1n s n<br />
2n<br />
<br />
1 n<br />
2<br />
∑ k1<br />
∑ n<br />
u<br />
k1 k<br />
2n<br />
→ 1 2 lim n→ u k<br />
<br />
→ 0asn → .<br />
u k 1 n<br />
2<br />
∑ k1<br />
−1 k s k<br />
n<br />
v n<br />
2n<br />
1 2 ∑ −1 n a n<br />
n1<br />
by <strong>The</strong>orem 8.48.<br />
(c) ∑ n1<br />
−1 n 1 1 2 ... 1 n − log 2 C,1.<br />
<br />
Proof: By (b) and ∑ n1<br />
Infinite products<br />
−1 n<br />
n<br />
− log 2, it is clear.<br />
8.39 Determine whether or not the following infinite products converges. Find the<br />
value of each convergent product.<br />
<br />
(a) n2<br />
1 −<br />
2<br />
nn1<br />
Proof: Consider<br />
we have<br />
1 − 2<br />
nn 1<br />
<br />
n − 1n 2<br />
nn 1<br />
,
n<br />
<br />
n2<br />
which implies that<br />
(b) <br />
n2<br />
1 − n −2 <br />
Proof: Consider<br />
we have<br />
which implies that<br />
<br />
(c) n2<br />
n 3 −1<br />
n 3 1<br />
Proof: Consider<br />
1 − 2<br />
kk 1<br />
n<br />
<br />
k2<br />
<br />
<br />
n2<br />
n<br />
<br />
n2<br />
k − 1k 2<br />
kk 1<br />
1 2 4<br />
3 2 3 5<br />
4 3 4 6<br />
5<br />
n 3n<br />
2<br />
1 − n −2 <br />
1 − 2<br />
nn 1<br />
1 − k −2 <br />
we have (let fk k − 1 2 k − 1 1),<br />
n<br />
<br />
k2<br />
1 3 .<br />
n − 1n 1<br />
nn ,<br />
n<br />
k2<br />
n 1<br />
2n<br />
k − 1k 1<br />
kk<br />
<br />
1 − n −2 1/2.<br />
n2<br />
n 3 − 1<br />
n 3 1 n − 1n2 n 1<br />
n 1n 2 − n 1<br />
n − 1n<br />
<br />
2 n 1<br />
n 1 n − 1 2 n − 1 1<br />
n<br />
k 3 − 1<br />
k 3 1 <br />
k2<br />
n<br />
− 1n 2<br />
nn 1<br />
k − 1k 2 k 1<br />
k 1 k − 1 2 k − 1 1<br />
2 3 n2 n 1<br />
nn 1<br />
which implies that<br />
<br />
n3 − 1<br />
n 3 1 2 3 .<br />
n2<br />
(d) n0<br />
1 z 2n <br />
if |z| 1.<br />
Proof: Consider<br />
n<br />
1 z 2k <br />
1 z1 z 2 1 z 2n <br />
<br />
k0
which implies that<br />
which implies that (if |z| 1)<br />
So,<br />
n<br />
<br />
k0<br />
n<br />
1 − z 1 z 2k <br />
1 − z 2n1 <br />
k0<br />
1 z 2k <br />
1 − z2n1 <br />
1 − z<br />
→ 1<br />
1 − z<br />
<br />
1 z 2n <br />
1<br />
1 − z .<br />
n0<br />
as n → .<br />
8.40 If each partial sum s n of the convergent series ∑ a n is not zero and if the sum<br />
itself is not zero, show that the infinite product a 1 n2<br />
1 − a n /s n−1 converges and has the<br />
value ∑ n1<br />
a n .<br />
Proof: Consider<br />
n<br />
n<br />
s<br />
a 1 1 a k /s k−1 a 1 k−1 a k<br />
s k−1<br />
k2<br />
k2<br />
n<br />
<br />
a 1 <br />
k2<br />
s k<br />
s k−1<br />
s n → ∑ a n ≠ 0.<br />
So, the infinite product a 1 <br />
n2<br />
1 − a n /s n−1 converges and has the value ∑ n1<br />
a n .<br />
8.41 Find the values of the following products by establishing the following identities<br />
and summing the series:<br />
(a) n2<br />
1 − 1 2 ∑ 2<br />
2 n −2 −n .<br />
n1<br />
Proof: Consider<br />
we have<br />
1 − 1<br />
2 n − 2 2n − 1<br />
2 n − 2 1 2 2n − 1<br />
2 n−1 − 1 ,<br />
n<br />
<br />
k2<br />
1 − 1<br />
2 k − 2<br />
n<br />
<br />
k2<br />
1<br />
2 2k − 1<br />
2 k−1 − 1<br />
n<br />
2 −n−1 <br />
k2<br />
2 k − 1<br />
2 k−1 − 1<br />
2 −n−1 2 n − 1<br />
2 −n−1 2 n−1 ...1<br />
1 ... 1<br />
n<br />
∑<br />
k1<br />
n<br />
2 ∑<br />
k1<br />
1<br />
2 k−1<br />
2 n−1<br />
1<br />
2 k .
So,<br />
<br />
(b) n2<br />
1 1<br />
n 2 −1<br />
Proof: Consider<br />
we have<br />
So,<br />
<br />
<br />
n2<br />
1 − 1<br />
2 n − 2<br />
1<br />
2 ∑ .<br />
n1 nn1<br />
1 1<br />
n 2 − 1 n 2<br />
n 2 − 1 <br />
n<br />
<br />
k2<br />
<br />
<br />
n2<br />
1 1<br />
k 2 − 1<br />
1 1<br />
n 2 − 1<br />
<br />
2 ∑ 2 −n<br />
n1<br />
2.<br />
nn<br />
n − 1n 1 ,<br />
n<br />
<br />
kk<br />
k − 1k 1<br />
k2<br />
2 n<br />
n 1<br />
2 1 − 1<br />
n<br />
2 ∑<br />
k1<br />
<br />
2 ∑<br />
n1<br />
2.<br />
n 1<br />
1<br />
kk 1 .<br />
1<br />
nn 1<br />
8.42 Determine all real x for which the product <br />
n1<br />
cosx/2 n converges and find the<br />
value of the product when it does converge.<br />
Proof: Ifx ≠ m, wherem ∈ Z, thensin x<br />
n<br />
<br />
k1<br />
cosx/2 k 2n sin x<br />
2 n<br />
2 n sin x<br />
n<br />
<br />
2 n k1<br />
2 n ≠ 0 for all n ∈ N. Hence,<br />
cosx/2 k <br />
sin x<br />
2 n sin x → sin x x .<br />
2 n<br />
If x m, wherem ∈ Z. <strong>The</strong>n as m 0, it is clear that the product converges to 1. So,<br />
we consider m ≠ 0 as follows. Since x m, choosing n large enough, i.e., as n ≥ N so<br />
that sin x ≠ 0. Hence,<br />
2 n<br />
and note that<br />
Hence,<br />
n<br />
<br />
k1<br />
N−1<br />
cosx/2 k <br />
k1<br />
lim n→<br />
n<br />
cosx/2 k cosx/2 k <br />
kN<br />
N−1<br />
cosx/2 k sinx/2<br />
<br />
N−1 <br />
2 n−N1 sinx/2 n <br />
k1<br />
sinx/2 N−1 <br />
2 n−N1 sinx/2 n sinx/2N−1 <br />
x/2 N−1 .<br />
<br />
cosx/2 k sinx/2N−1 N−1<br />
<br />
cosx/2<br />
x/2 N−1 k .<br />
k1<br />
k1
So, by above sayings, we have prove that the convergence of the product for all x ∈ R.<br />
8.43 (a) Let a n −1 n / n for n 1,2,... Show that1 a n diverges but that<br />
∑ a n converges.<br />
Proof: Clearly, ∑ a n converges since it is alternating series. Consider<br />
2n<br />
<br />
k2<br />
2n<br />
1 a k 1 −1k<br />
k2<br />
k<br />
1 1 2<br />
≤ 1 1 2<br />
1 − 1 3<br />
1 − 1 4<br />
1 1 4<br />
1 1 4<br />
1 − 1<br />
2n − 1<br />
1 − 1<br />
2n<br />
1 1<br />
2n<br />
1 1<br />
2n<br />
1 1 2<br />
1 − 1 4<br />
1 −<br />
1 2n<br />
*<br />
and note that<br />
n<br />
1 −<br />
2k 1 : p n<br />
k2<br />
is decreasing. From the divergence of ∑ 1 , we know that p 2k n → 0. So,<br />
That is, <br />
k2<br />
1 a k diverges to zero.<br />
<br />
1 a k 0.<br />
k2<br />
(b) Let a 2n−1 −1/ n , a 2n 1/ n 1/n for n 1, 2, . . . Show that 1 a n <br />
converges but ∑ a n diverges.<br />
Proof: Clearly, ∑ a n diverges. Consider<br />
and<br />
2n<br />
1 a k 1 a 2 1 a 3 1 a 4 1 a 2n <br />
k2<br />
31 a 3 1 a 4 1 a 2n <br />
3 1 − 1<br />
2 2<br />
1 − 1<br />
n n<br />
2n1<br />
1 a k 1 a 2 1 a 3 1 a 4 1 a 2n 1 a 2n1 <br />
k2<br />
3 1 − 1 1 − 1<br />
2 2<br />
n n<br />
By (*) and (**), we know that<br />
1 a n converges<br />
n<br />
since k2<br />
1 − 1<br />
k k<br />
converges.<br />
1 − 1<br />
n 1<br />
8.44 Assume that a n ≥ 0 for each n 1, 2, . . . Assume further that<br />
*<br />
**
Show that k1<br />
a 2n2<br />
1 a 2n2<br />
a 2n1 <br />
a 2n<br />
1 a 2n<br />
for n 1, 2, . . .<br />
1 −1 k a k converges if, and only if, ∑ k1<br />
−1 k a k converges.<br />
a<br />
1c<br />
Proof: First, we note that if b, then 1 a1 − b 1, and if b <br />
1a<br />
c , then<br />
1 1 − b1 c. Hence, by hypothesis, we have<br />
1 1 a 2n 1 − a 2n1 *<br />
and<br />
1 1 a 2n2 1 − a 2n1 . **<br />
()Suppose that ∑ <br />
k1<br />
−1 k a k converges, then lim k→ a k 0. Consider Cauchy<br />
Condition for product,<br />
1 −1 p1 a p1 1 −1 p2 a p2 1 −1 pq a pq − 1 for q 1, 2, 3, . . . .<br />
If p 1 2m, andq 2l, then<br />
1 −1 p1 a p1 1 −1 p2 a p2 1 −1 pq a pq − 1<br />
|1 a 2m 1 − a 2m1 1 a 2m2l − 1|<br />
≤ 1 a 2m − 1by(*)and(**)<br />
a 2m → 0.<br />
<br />
Similarly for other cases, so we have proved that k1<br />
1 −1 k a k converges by<br />
Cauchy Condition for product.<br />
()This is a counterexample as follows. Let a n −1 n<br />
n, thenitiseasytoshowthat<br />
a 2n2<br />
a<br />
1 a 2n1 <br />
2n2<br />
In addition,<br />
n<br />
<br />
k1<br />
1 −1 k a k <br />
k1<br />
However, consider<br />
n<br />
∑<br />
k1<br />
n<br />
∑<br />
k1<br />
n<br />
∑<br />
k1<br />
n<br />
n<br />
a 2k − a 2k−1 <br />
exp 1<br />
2k<br />
exp −1k<br />
k<br />
− exp<br />
exp<br />
expb k 1<br />
2k 1<br />
2k − 1<br />
<br />
exp −1n<br />
n<br />
a 2n<br />
1 a 2n<br />
for n 1, 2, . . .<br />
−1<br />
2k − 1<br />
n<br />
∑<br />
k1<br />
−1 k<br />
k<br />
,whereb k ∈<br />
≥ ∑ exp−1 1<br />
k1<br />
2k 1<br />
2k − 1<br />
So, by <strong>The</strong>orem 8.13, we proved the divergence of ∑ k1<br />
→ as n → .<br />
<br />
−1 k a k .<br />
− 1<br />
≥ 0 for all<br />
→ exp− log 2 as n → .<br />
−1<br />
2k − 1 , 1<br />
2k<br />
8.45 A complex-valued sequence fn is called multiplicative if f1 1andif<br />
fmn fmfn whenever m and n are relatively prime. (See Section 1.7) It is called<br />
completely multiplicative if<br />
f1 1andiffmn fmfn for all m and n.
(a) If fn is multiplicative and if the series ∑ fn converges absolutely, prove that<br />
<br />
∑<br />
n1<br />
fn 1 fp k fp k2 ...,<br />
k1<br />
where p k denote the kth prime, the product being absolutely convergent.<br />
<br />
Proof: We consider the partial product P m m<br />
k1<br />
1 fp k fp k2 ... and show<br />
that P m → ∑ <br />
n1<br />
fn as m → . Writing each factor as a geometric series we have<br />
m<br />
P m 1 fp k fp k2 ...,<br />
k1<br />
a product of a finite number of absolutely convergent series. When we multiple these series<br />
together and rearrange the terms such that a typical term of the new absolutely convergent<br />
series is<br />
fn fp 1<br />
a 1<br />
fp m<br />
a m<br />
,wheren p 1<br />
a 1<br />
p m<br />
a m<br />
,<br />
and each a i ≥ 0. <strong>The</strong>refore, we have<br />
P m ∑ fn,<br />
1<br />
where ∑ 1<br />
is summed over those n having all their prime factors ≤ p m .Bytheunique<br />
factorization theorem (<strong>The</strong>orem 1.9), each such n occors once and only once in ∑ 1<br />
.<br />
Substracting P m from ∑ <br />
n1<br />
fn, weget<br />
<br />
<br />
∑ fn − P m ∑ fn − ∑ fn ∑ fn<br />
n1<br />
n1<br />
1<br />
2<br />
where ∑ 2<br />
is summed over those n having at least one prime factor p m . Since these n<br />
occors among the integers p m ,wehave<br />
<br />
∑<br />
n1<br />
fn − P m<br />
<br />
≤ ∑<br />
np m<br />
|fn|.<br />
As m → the last sum tends to 0 because ∑ n1<br />
fn converges, so P m → ∑ n1<br />
fn.<br />
To prove that the product converges absolutely we use <strong>The</strong>orem 8.52. <strong>The</strong> product has<br />
the form 1 a k ,where<br />
a k fp k fp k2 ....<br />
<br />
<strong>The</strong> series ∑|a k | converges since it is dominated by ∑ n1|fn|.<br />
<strong>The</strong>reofore, 1 ak <br />
also converges absolutely.<br />
Remark: <strong>The</strong> method comes from Euler. By the same method, it also shows that there<br />
are infinitely many primes. <strong>The</strong> reader can see the book, An Introduction To <strong>The</strong> <strong>The</strong>ory<br />
Of <strong>Number</strong>s by Loo-Keng Hua, pp 91-93. (Chinese Version)<br />
(b) If, in addition, fn is completely multiplicative, prove that the formula in (a)<br />
becomes<br />
<br />
<br />
∑ fn <br />
1<br />
1 − fp<br />
n1<br />
k1<br />
k .<br />
Note that Euler’s product for s (<strong>The</strong>orem 8.56) is the special case in which fn n −s .<br />
Proof: By(a),iffn is completely multiplicative, then rewrite
1 fp k fp k2 ... ∑fp k n<br />
n0<br />
1<br />
1 − fp k <br />
since |fp k | 1 for all p k . (Suppose NOT, then |fp k | ≥ 1 |fp kn | |fp k | n ≥ 1<br />
contradicts to lim n→ fn 0. ).<br />
Hence,<br />
<br />
∑<br />
n1<br />
<br />
fn <br />
k1<br />
1<br />
1 − fp k .<br />
8.46 This exercise outlines a simple proof of the formula 2 2 /6. Start with the<br />
inequality sinx x tan x, valid for 0 x /2, taking recipocals, and square each<br />
member to obtain<br />
cot 2 x 1 1 cot<br />
x 2 x.<br />
2<br />
Now put x k/2m 1, wherek and m are integers, with 1 ≤ k ≤ m, and sum on k to<br />
obtain<br />
m<br />
∑ cot 2 k<br />
m<br />
m<br />
2m 12<br />
<br />
2m 1<br />
∑<br />
1 m ∑ cot<br />
2 k 2<br />
2<br />
2m k<br />
1 .<br />
k1<br />
k1<br />
k1<br />
Use the formula of Exercise 1.49(c) to deduce the ineqaulity<br />
m2m − 1 2<br />
32m 1 2<br />
m<br />
∑<br />
k1<br />
1<br />
k 2<br />
<br />
2mm 12<br />
32m 1 2<br />
Now let m → to obtain 2 2 /6.<br />
Proof: <strong>The</strong> proof is clear if we follow the hint and Exercise 1.49 (c), soweomitit.<br />
8.47 Use an argument similar to that outlined in Exercise 8.46 to prove that<br />
4 4 /90.<br />
Proof: <strong>The</strong> proof is clear if we follow the Exercise 8.46 and Exercise 1.49 (c), sowe<br />
omit it.<br />
Remark: (1) From this, it is easy to compute the value of 2s, where<br />
s ∈ n : n ∈ N. In addition, we will learn some new method such as Fourier series and<br />
so on, to find the value of Riemann zeta function.<br />
(2) <strong>The</strong>r is an open problem that 2s − 1, wheres ∈ n ∈ N : n 1.
Sequences of Functions<br />
Uniform convergence<br />
9.1 Assume that f n → f uniformly on S and that each f n is bounded on<br />
S. Prove that {f n } is uniformly bounded on S.<br />
Proof: Since f n → f uniformly on S, then given ε = 1, there exists a<br />
positive integer n 0 such that as n ≥ n 0 , we have<br />
|f n (x) − f (x)| ≤ 1 for all x ∈ S. (*)<br />
Hence, f (x) is bounded on S by the following<br />
|f (x)| ≤ |f n0 (x)| + 1 ≤ M (n 0 ) + 1 for all x ∈ S. (**)<br />
where |f n0 (x)| ≤ M (n 0 ) for all x ∈ S.<br />
Let |f 1 (x)| ≤ M (1) , ..., |f n0 −1 (x)| ≤ M (n 0 − 1) for all x ∈ S, then by<br />
(*) and (**),<br />
So,<br />
|f n (x)| ≤ 1 + |f (x)| ≤ M (n 0 ) + 2 for all n ≥ n 0 .<br />
|f n (x)| ≤ M for all x ∈ S and for all n<br />
where M = max (M (1) , ..., M (n 0 − 1) , M (n 0 ) + 2) .<br />
Remark: (1) In the proof, we also shows that the limit function f is<br />
bounded on S.<br />
(2) <strong>The</strong>re is another proof. We give it as a reference.<br />
Proof: Since Since f n → f uniformly on S, then given ε = 1, there exists<br />
a positive integer n 0 such that as n ≥ n 0 , we have<br />
|f n (x) − f n+k (x)| ≤ 1 for all x ∈ S and k = 1, 2, ...<br />
So, for all x ∈ S, and k = 1, 2, ...<br />
|f n0 +k (x)| ≤ 1 + |f n0 (x)| ≤ M (n 0 ) + 1 (*)<br />
where |f n0 (x)| ≤ M (n 0 ) for all x ∈ S.<br />
Let |f 1 (x)| ≤ M (1) , ..., |f n0 −1 (x)| ≤ M (n 0 − 1) for all x ∈ S, then by<br />
(*),<br />
|f n (x)| ≤ M for all x ∈ S and for all n<br />
1
where M = max (M (1) , ..., M (n 0 − 1) , M (n 0 ) + 1) .<br />
9.2 Define two sequences {f n } and {g n } as follows:<br />
(<br />
f n (x) = x 1 + 1 )<br />
if x ∈ R, n = 1, 2, ...,<br />
n<br />
g n (x) =<br />
{ 1<br />
n<br />
Let h n (x) = f n (x) g n (x) .<br />
if x = 0 or if x is irrational,<br />
b + 1 n if x is rational, say x = a b , b > 0.<br />
(a) Prove that both {f n } and {g n } converges uniformly on every bounded<br />
interval.<br />
and<br />
Proof: Note that it is clear that<br />
lim f n (x) = f (x) = x, for all x ∈ R<br />
n→∞<br />
{ 0 if x = 0 or if x is irrational,<br />
lim g n (x) = g (x) =<br />
n→∞ b if x is ratonal, say x = a, b > 0.<br />
b<br />
In addition, in order to show that {f n } and {g n } converges uniformly<br />
on every bounded interval, it suffices to consider the case of any compact<br />
interval [−M, M] , M > 0.<br />
Given ε > 0, there exists a positive integer N such that as n ≥ N, we<br />
have<br />
M<br />
n < ε and 1 n < ε.<br />
Hence, for this ε, we have as n ≥ N<br />
|f n (x) − f (x)| = ∣ x ∣ ≤ M < ε for all x ∈ [−M, M]<br />
n n<br />
and<br />
|g n (x) − g (x)| ≤ 1 n<br />
< ε for all x ∈ [−M, M] .<br />
That is, we have proved that {f n } and {g n } converges uniformly on every<br />
bounded interval.<br />
Remark: In the proof, we use the easy result directly from definition<br />
of uniform convergence as follows. If f n → f uniformly on S, then f n → f<br />
uniformly on T for every subset T of S.<br />
2
(b) Prove that h n (x) does not converges uniformly on any bounded interval.<br />
Proof: Write<br />
{<br />
h n (x) =<br />
a + a n<br />
x<br />
n<br />
(<br />
1 +<br />
1<br />
( n)<br />
if x = 0 or x is irrational<br />
1 +<br />
1<br />
+ ) 1<br />
b bn if x is rational, say x =<br />
a<br />
b<br />
<strong>The</strong>n<br />
{ 0 if x = 0 or x is irrational<br />
lim h n (x) = h (x) =<br />
n→∞ a if x is rational, say x = a .<br />
b<br />
Hence, if h n (x) converges uniformly on any bounded interval I, then h n (x)<br />
converges uniformly on [c, d] ⊆ I. So, given ε = max (|c| , |d|) > 0, there is a<br />
positive integer N such that as n ≥ N, we have<br />
max (|c| , |d|) > |h n (x) − h (x)|<br />
{ ∣ ( )∣ x<br />
= n 1 +<br />
1 ∣<br />
n =<br />
|x| ∣<br />
n 1 +<br />
1 ∣ ( n if x ∈ Q c ∩ [c, d] or x = 0<br />
∣ a<br />
n 1 +<br />
1<br />
+ )∣ 1 ∣<br />
b bn if x ∈ Q ∩ [c, d] , x =<br />
a<br />
b<br />
which implies that (x ∈ [c, d] ∩ Q c or x = 0)<br />
max (|c| , |d|) > |x|<br />
n ∣ 1 + 1 n∣ ≥ |x|<br />
n<br />
≥<br />
max (|c| , |d|)<br />
n<br />
which is absurb. So, h n (x) does not converges uniformly on any bounded<br />
interval.<br />
9.3 Assume that f n → f uniformly on S, g n → f uniformly on S.<br />
(a) Prove that f n + g n → f + g uniformly on S.<br />
Proof: Since f n → f uniformly on S, and g n → f uniformly on S, then<br />
given ε > 0, there is a positive integer N such that as n ≥ N, we have<br />
.<br />
and<br />
|f n (x) − f (x)| < ε 2<br />
|g n (x) − g (x)| < ε 2<br />
for all x ∈ S<br />
for all x ∈ S.<br />
Hence, for this ε, we have as n ≥ N,<br />
|f n (x) + g n (x) − f (x) − g (x)| ≤ |f n (x) − f (x)| + |g n (x) − g (x)|<br />
< ε for all x ∈ S.<br />
3
That is, f n + g n → f + g uniformly on S.<br />
Remark: <strong>The</strong>re is a similar result. We write it as follows. If f n → f<br />
uniformly on S, then cf n → cf uniformly on S for any real c. Since the proof<br />
is easy, we omit the proof.<br />
(b) Let h n (x) = f n (x) g n (x) , h (x) = f (x) g (x) , if x ∈ S. Exercise 9.2<br />
shows that the assertion h n → h uniformly on S is, in general, incorrect.<br />
Prove that it is correct if each f n and each g n is bounded on S.<br />
Proof: Since f n → f uniformly on S and each f n is bounded on S, then<br />
f is bounded on S by Remark (1) in the Exercise 9.1. In addition, since<br />
g n → g uniformly on S and each g n is bounded on S, then g n is uniformly<br />
bounded on S by Exercise 9.1.<br />
Say |f (x)| ≤ M 1 for all x ∈ S, and |g n (x)| ≤ M 2 for all x and all n. <strong>The</strong>n<br />
given ε > 0, there exists a positive integer N such that as n ≥ N, we have<br />
and<br />
|f n (x) − f (x)| <<br />
|g n (x) − g (x)| <<br />
which implies that as n ≥ N, we have<br />
ε<br />
2 (M 2 + 1)<br />
ε<br />
2 (M 1 + 1)<br />
|h n (x) − h (x)| = |f n (x) g n (x) − f (x) g (x)|<br />
for all x ∈ S<br />
for all x ∈ S<br />
= |[f n (x) − f (x)] [g n (x)] + [f (x)] [g n (x) − g (x)]|<br />
≤ |f n (x) − f (x)| |g n (x)| + |f (x)| |g n (x) − g (x)|<br />
ε<br />
<<br />
2 (M 2 + 1) M ε<br />
2 + M 1<br />
2 (M 1 + 1)<br />
< ε 2 + ε 2<br />
= ε<br />
for all x ∈ S. So, h n → h uniformly on S.<br />
9.4 Assume that f n → f uniformly on S and suppose there is a constant<br />
M > 0 such that |f n (x)| ≤ M for all x in S and all n. Let g be continuous<br />
on the closure of the disk B (0; M) and define h n (x) = g [f n (x)] , h (x) =<br />
g [f (x)] , if x ∈ S. Prove that h n → h uniformly on S.<br />
4
Proof: Since g is continuous on a compact disk B (0; M) , g is uniformly<br />
continuous on B (0; M) . Given ε > 0, there exists a δ > 0 such that as<br />
|x − y| < δ, where x, y ∈ S, we have<br />
|g (x) − g (y)| < ε. (*)<br />
For this δ > 0, since f n → f uniformly on S, then there exists a positive<br />
integer N such that as n ≥ N, we have<br />
|f n (x) − f (x)| < δ for all x ∈ S. (**)<br />
Hence, by (*) and (**), we conclude that given ε > 0, there exists a positive<br />
integer N such that as n ≥ N, we have<br />
Hence, h n → h uniformly on S.<br />
|g (f n (x)) − g (f (x))| < ε for all x ∈ S.<br />
9.5 (a) Let f n (x) = 1/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that {f n }<br />
converges pointwise but not uniformly on (0, 1) .<br />
Proof: First, it is clear that lim n→∞ f n (x) = 0 for all x ∈ (0, 1) . Supppos<br />
that {f n } converges uniformly on (0, 1) . <strong>The</strong>n given ε = 1/2, there exists a<br />
positive integer N such that as n ≥ N, we have<br />
|f n (x) − f (x)| =<br />
1<br />
∣1 + nx∣ < 1/2 for all x ∈ (0, 1) .<br />
So, the inequality holds for all x ∈ (0, 1) . It leads us to get a contradiction<br />
since<br />
1<br />
1 + Nx < 1 2<br />
for all x ∈ (0, 1) ⇒ lim<br />
x→0 + 1<br />
1 + Nx = 1 < 1/2.<br />
That is, {f n } converges NOT uniformly on (0, 1) .<br />
(b) Let g n (x) = x/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that g n → 0<br />
uniformly on (0, 1) .<br />
Proof: First, it is clear that lim n→∞ g n (x) = 0 for all x ∈ (0, 1) . Given<br />
ε > 0, there exists a positive integer N such that as n ≥ N, we have<br />
1/n < ε<br />
5
which implies that<br />
∣ ∣ |g n (x) − g| =<br />
x<br />
∣∣∣ ∣1 + nx∣ = 1 ∣∣∣<br />
1<br />
+ n < 1 n < ε.<br />
x<br />
So, g n → 0 uniformly on (0, 1) .<br />
9.6 Let f n (x) = x n . <strong>The</strong> sequence {f n (x)} converges pointwise but not<br />
uniformly on [0, 1] . Let g be continuous on [0, 1] with g (1) = 0. Prove that<br />
the sequence {g (x) x n } converges uniformly on [0, 1] .<br />
Proof: It is clear that f n (x) = x n converges NOT uniformly on [0, 1]<br />
since each term of {f n (x)} is continuous on [0, 1] and its limit function<br />
{ 0 if x ∈ [0, 1)<br />
f =<br />
1 if x = 1.<br />
is not a continuous function on [0, 1] by <strong>The</strong>orem 9.2.<br />
In order to show {g (x) x n } converges uniformly on [0, 1] , it suffices to<br />
shows that {g (x) x n } converges uniformly on [0, 1). Note that<br />
lim g (x)<br />
n→∞ xn = 0 for all x ∈ [0, 1).<br />
We partition the interval [0, 1) into two subintervals: [0, 1 − δ] and (1 − δ, 1) .<br />
As x ∈ [0, 1 − δ] : Let M = max x∈[0,1] |g (x)| , then given ε > 0, there is a<br />
positive integer N such that as n ≥ N, we have<br />
M (1 − δ) n < ε<br />
which implies that for all x ∈ [0, 1 − δ] ,<br />
|g (x) x n − 0| ≤ M |x n | ≤ M (1 − δ) n < ε.<br />
Hence, {g (x) x n } converges uniformly on [0, 1 − δ] .<br />
As x ∈ (1 − δ, 1) : Since g is continuous at 1, given ε > 0, there exists a<br />
δ > 0 such that as |x − 1| < δ, where x ∈ [0, 1] , we have<br />
|g (x) − g (1)| = |g (x) − 0| = |g (x)| < ε<br />
which implies that for all x ∈ (1 − δ, 1) ,<br />
|g (x) x n − 0| ≤ |g (x)| < ε.<br />
6
Hence, {g (x) x n } converges uniformly on (1 − δ, 1) .<br />
So, from above sayings, we have proved that the sequence of functions<br />
{g (x) x n } converges uniformly on [0, 1] .<br />
Remark: It is easy to show the followings by definition. So, we omit the<br />
proof.<br />
(1) Suppose that for all x ∈ S, the limit function f exists. If f n → f<br />
uniformly on S 1 (⊆ S) , then f n → f uniformly on S, where # (S − S 1 ) <<br />
+∞.<br />
(2) Suppose that f n → f uniformly on S and on T. <strong>The</strong>n f n → f uniformly<br />
on S ∪ T.<br />
9.7 Assume that f n → f uniformly on S and each f n is continuous on S.<br />
If x ∈ S, let {x n } be a sequence of points in S such that x n → x. Prove that<br />
f n (x n ) → f (x) .<br />
Proof: Since f n → f uniformly on S and each f n is continuous on S, by<br />
<strong>The</strong>orem 9.2, the limit function f is also continuous on S. So, given ε > 0,<br />
there is a δ > 0 such that as |y − x| < δ, where y ∈ S, we have<br />
|f (y) − f (x)| < ε 2 .<br />
For this δ > 0, there exists a positive integer N 1 such that as n ≥ N 1 , we<br />
have<br />
|x n − x| < δ.<br />
Hence, as n ≥ N 1 , we have<br />
|f (x n ) − f (x)| < ε 2 . (*)<br />
In addition, since f n → f uniformly on S, given ε > 0, there exists a<br />
positive integer N ≥ N 1 such that as n ≥ N, we have<br />
|f n (x) − f (x)| < ε 2<br />
for all x ∈ S<br />
which implies that<br />
|f n (x n ) − f (x n )| < ε 2 . (**)<br />
7
By (*) and (**), we obtain that given ε > 0, there exists a positie integer N<br />
such that as n ≥ N, we have<br />
|f n (x n ) − f (x)| = |f n (x n ) − f (x n )| + |f (x n ) − f (x)|<br />
< ε 2 + ε 2<br />
= ε.<br />
That is, we have proved that f n (x n ) → f (x) .<br />
9.8 Let {f n } be a seuqnece of continuous functions defined on a compact<br />
set S and assume that {f n } converges pointwise on S to a limit function f.<br />
Prove that f n → f uniformly on S if, and only if, the following two conditions<br />
hold.:<br />
(i) <strong>The</strong> limit function f is continuous on S.<br />
(ii) For every ε > 0, there exists an m > 0 and a δ > 0, such that n > m<br />
and |f k (x) − f (x)| < δ implies |f k+n (x) − f (x)| < ε for all x in S and all<br />
k = 1, 2, ...<br />
Hint. To prove the sufficiency of (i) and (ii), show that for each x 0 in S<br />
there is a neighborhood of B (x 0 ) and an integer k (depending on x 0 ) such<br />
that<br />
|f k (x) − f (x)| < δ if x ∈ B (x 0 ) .<br />
By compactness, a finite set of integers, say A = {k 1 , ..., k r } , has the property<br />
that, for each x in S, some k in A satisfies |f k (x) − f (x)| < δ. Uniform<br />
convergence is an easy consequences of this fact.<br />
Proof: (⇒) Suppose that f n → f uniformly on S, then by <strong>The</strong>orem<br />
9.2, the limit function f is continuous on S. In addition, given ε > 0, there<br />
exists a positive integer N such that as n ≥ N, we have<br />
|f n (x) − f (x)| < ε for all x ∈ S<br />
Let m = N, and δ = ε, then (ii) holds.<br />
(⇐) Suppose that (i) and (ii) holds. We prove f k → f uniformly on S as<br />
follows. By (ii), given ε > 0, there exists an m > 0 and a δ > 0, such that<br />
n > m and |f k (x) − f (x)| < δ implies |f k+n (x) − f (x)| < ε for all x in S<br />
and all k = 1, 2, ...<br />
Consider ∣ ∣f k(x0 ) (x 0 ) − f (x 0 ) ∣ ∣ < δ, then there exists a B (x 0 ) such that as<br />
x ∈ B (x 0 ) ∩ S, we have<br />
∣ fk(x0 ) (x) − f (x) ∣ ∣ < δ<br />
8
y continuity of f k(x0 ) (x) − f (x) . Hence, by (ii) as n > m<br />
∣<br />
∣f k(x0 )+n (x) − f (x) ∣ ∣ < ε if x ∈ B (x 0 ) ∩ S. (*)<br />
Note that S is compact and S = ∪ x∈S (B (x) ∩ S) , then S = ∪ p k=1 (B (x k) ∩ S) .<br />
So, let N = max p i=1 (k (x p) + m) , as n > N, we have<br />
|f n (x) − f (x)| < ε for all x ∈ S<br />
with help of (*). That is, f n → f uniformly on S.<br />
9.9 (a) Use Exercise 9.8 to prove the following theorem of Dini: If<br />
{f n } is a sequence of real-valued continuous functions converginf<br />
pointwise to a continuous limit function f on a compact set S, and<br />
if f n (x) ≥ f n+1 (x) for each x in S and every n = 1, 2, ..., then f n → f<br />
uniformly on S.<br />
Proof: By Exercise 9.8, in order to show that f n → f uniformly on S,<br />
it suffices to show that (ii) holds. Since f n (x) → f (x) and f n+1 (x) ≤ f n (x)<br />
on S, then fixed x ∈ S, and given ε > 0, there exists a positive integer<br />
N (x) = N such that as n ≥ N, we have<br />
0 ≤ f n (x) − f (x) < ε.<br />
Choose m = 1 and δ = ε, then by f n+1 (x) ≤ f n (x) , then (ii) holds. We<br />
complete it.<br />
Remark: (1) Dini’s <strong>The</strong>orem is important in Analysis; we suggest the<br />
reader to keep it in mind.<br />
(2) <strong>The</strong>re is another proof by using Cantor Intersection <strong>The</strong>orem.<br />
We give it as follows.<br />
Proof: Let g n = f n − f, then g n is continuous on S, g n → 0 pointwise on<br />
S, and g n (x) ≥ g n+1 (x) on S. If we can show g n → 0 uniformly on S, then<br />
we have proved that f n → f uniformly on S.<br />
Given ε > 0, and consider S n := {x : g n (x) ≥ ε} . Since each g n (x) is<br />
continuous on a compact set S, we obtain that S n is compact. In addition,<br />
S n+1 ⊆ S n since g n (x) ≥ g n+1 (x) on S. <strong>The</strong>n<br />
∩S n ≠ φ (*)<br />
9
if each S n is non-empty by Cantor Intersection <strong>The</strong>orem. However (*)<br />
contradicts to g n → 0 pointwise on S. Hence, we know that there exists a<br />
positive integer N such that as n ≥ N,<br />
S n = φ.<br />
That is, given ε > 0, there exists a positive integer N such that as n ≥ N,<br />
we have<br />
|g n (x) − 0| < ε.<br />
So, g n → 0 uniformly on S.<br />
(b) Use he sequence in Exercise 9.5(a) to show that compactness of S is<br />
essential in Dini’s <strong>The</strong>orem.<br />
Proof: Let f n (x) = 1 , where x ∈ (0, 1) . <strong>The</strong>n it is clear that each<br />
1+nx<br />
f n (x) is continuous on (0, 1) , the limit function f (x) = 0 is continuous on<br />
(0, 1) , and f n+1 (x) ≤ f n (x) for all x ∈ (0, 1) . However, f n → f not uniformly<br />
on (0, 1) by Exercise 9.5 (a). Hence, compactness of S is essential in Dini’s<br />
<strong>The</strong>orem.<br />
9.10 Let f n (x) = n c x (1 − x 2 ) n for x real and n ≥ 1. Prove that {f n }<br />
converges pointwsie on [0, 1] for every real c. Determine those c for which the<br />
convergence is uniform on [0, 1] and those for which term-by-term integration<br />
on [0, 1] leads to a correct result.<br />
Proof: It is clear that f n (0) → 0 and f n (1) → 0. Consider x ∈ (0, 1) ,<br />
then |1 − x 2 | := r < 1, then<br />
lim f n (x) = lim n c r n x = 0 for any real c.<br />
n→∞ n→∞<br />
Hence, f n → 0 pointwise on [0, 1] .<br />
Consider<br />
f n ′ (x) = n ( c 1 − x 2) ( )<br />
n−1 1<br />
(2n − 1)<br />
2n − 1 − x2 ,<br />
then each f n has the absolute maximum at x n = 1 √ 2n−1<br />
.<br />
As c < 1/2, we obtain that<br />
|f n (x)| ≤ |f n (x n )|<br />
(<br />
n c<br />
= √ 1 − 1 ) n<br />
2n − 1 2n − 1<br />
[√ (<br />
= n c− 1 n<br />
2<br />
1 − 1 ) n ]<br />
2n − 1 2n − 1<br />
10<br />
→ 0 as n → ∞. (*)
In addition, as c ≥ 1/2, if f n → 0 uniformly on [0, 1] , then given ε > 0, there<br />
exists a positive integer N such that as n ≥ N, we have<br />
which implies that as n ≥ N,<br />
which contradicts to<br />
|f n (x)| < ε for all x ∈ [0, 1]<br />
|f n (x n )| < ε<br />
{<br />
lim f √ 1<br />
n (x n ) = 2e<br />
if c = 1/2<br />
n→∞ ∞ if c > 1/2 . (**)<br />
From (*) and (**), we conclude that only as c < 1/2, the seqences of<br />
functions converges uniformly on [0, 1] .<br />
In order to determine those c for which term-by-term integration on [0, 1] ,<br />
we consider ∫ 1<br />
n c<br />
f n (x) dx =<br />
2 (n + 1)<br />
and ∫ 1<br />
f (x) dx =<br />
0<br />
0<br />
∫ 1<br />
0<br />
0dx = 0.<br />
Hence, only as c < 1, we can integrate it term-by-term.<br />
9.11 Prove that ∑ x n (1 − x) converges pointwise but not uniformly on<br />
[0, 1] , whereas ∑ (−1) n x n (1 − x) converges uniformly on [0, 1] . This illustrates<br />
that uniform convergence of ∑ f n (x) along with pointwise convergence<br />
of ∑ |f n (x)| does not necessarily imply uniform convergence<br />
of ∑ |f n (x)| .<br />
Proof: Let s n (x) = ∑ n<br />
k=0 xk (1 − x) = 1 − x n+1 , then<br />
{ 1 if x ∈ [0, 1)<br />
s n (x) →<br />
.<br />
0 if x = 1<br />
Hence, ∑ x n (1 − x) converges pointwise but not uniformly on [0, 1] by <strong>The</strong>orem<br />
9.2 since each s n is continuous on [0, 1] .<br />
Let g n (x) = x n (1 − x) , then it is clear that g n (x) ≥ g n+1 (x) for all x ∈<br />
[0, 1] , and g n (x) → 0 uniformly on [0, 1] by Exercise 9.6. Hence, by Dirichlet’s<br />
Test for uniform convergence, we have proved that ∑ (−1) n x n (1 − x)<br />
converges uniformly on [0, 1] .<br />
11
9.12 Assume that g n+1 (x) ≤ g n (x) for each x in T and each n = 1, 2, ...,<br />
and suppose that g n → 0 uniformly on T. Prove that ∑ (−1) n+1 g n (x) converges<br />
uniformly on T.<br />
Proof: It is clear by Dirichlet’s Test for uniform convergence.<br />
9.13 Prove Abel’s test for uniform convergence: Let {g n } be a sequence<br />
of real-valued functions such that g n+1 (x) ≤ g n (x) for each x in T and for<br />
every n = 1, 2, ... If {g n } is uniformly bounded on T and if ∑ f n (x) converges<br />
uniformly on T, then ∑ f n (x) g n (x) also converges uniformly on T.<br />
Proof: Let F n (x) = ∑ n<br />
k=1 f k (x) . <strong>The</strong>n<br />
s n (x) =<br />
n∑<br />
f k (x) g k (x) = F n g 1 (x)+<br />
k=1<br />
and hence if n > m, we can write<br />
n∑<br />
(F n (x) − F k (x)) (g k+1 (x) − g k (x))<br />
k=1<br />
s n (x)−s m (x) = (F n (x) − F m (x)) g m+1 (x)+<br />
n∑<br />
k=m+1<br />
Hence, if M is an uniform bound for {g n } , we have<br />
|s n (x) − s m (x)| ≤ M |F n (x) − F m (x)| + 2M<br />
(F n (x) − F k (x)) (g k+1 (x) − g k (x))<br />
n∑<br />
k=m+1<br />
|F n (x) − F k (x)| . (*)<br />
Since ∑ f n (x) converges uniformly on T, given ε > 0, there exists a positive<br />
integer N such that as n > m ≥ N, we have<br />
|F n (x) − F m (x)| < ε<br />
M + 1<br />
By (*) and (**), we have proved that as n > m ≥ N,<br />
|s n (x) − s m (x)| < ε for all x ∈ T.<br />
Hence, ∑ f n (x) g n (x) also converges uniformly on T.<br />
for all x ∈ T (**)<br />
Remark: In the proof, we establish the lemma as follows. We write it<br />
as a reference.<br />
12
(Lemma) If {a n } and {b n } are two sequences of complex numbers, define<br />
A n =<br />
n∑<br />
a k .<br />
k=1<br />
<strong>The</strong>n we have the identity<br />
n∑<br />
n∑<br />
a k b k = A n b n+1 − A k (b k+1 − b k )<br />
(i)<br />
k=1<br />
= A n b 1 +<br />
k=1<br />
n∑<br />
(A n − A k ) (b k+1 − b k ) . (ii)<br />
k=1<br />
Proof: <strong>The</strong> identity (i) comes from <strong>The</strong>orem 8.27. In order to show<br />
(ii), it suffices to consider<br />
b n+1 = b 1 +<br />
n∑<br />
b k+1 − b k .<br />
k=1<br />
9.14 Let f n (x) = x/ (1 + nx 2 ) if x ∈ R, n = 1, 2, ... Find the limit function<br />
f of the sequence {f n } and the limit function g of the sequence {f ′ n} .<br />
(a) Prove that f ′ (x) exists for every x but that f ′ (0) ≠ g (0) . For what<br />
values of x is f ′ (x) = g (x)<br />
Proof: It is easy to show that the limit function f = 0, and by f ′ n (x) =<br />
1−nx 2<br />
(1+nx 2 ) 2 , we have<br />
{ 1 if x = 0<br />
lim f n ′ (x) = g (x) =<br />
n→∞ 0 if x ≠ 0 .<br />
Hence, f ′ (x) exists for every x and f ′ (0) = 0 ≠ g (0) = 1. In addition, it is<br />
clear that as x ≠ 0, we have f ′ (x) = g (x) .<br />
(b) In what subintervals of R does f n → f uniformly<br />
Proof: Note that<br />
1 + nx 2<br />
2<br />
≥ √ n |x|<br />
13
y A.P. ≥ G.P. for all real x. Hence,<br />
∣ x ∣∣∣<br />
∣ ≤ 1<br />
1 + nx 2 2 √ n<br />
which implies that f n → f uniformly on R.<br />
(c) In what subintervals of R does f ′ n → g uniformly<br />
Proof: Since each f n ′ =<br />
1−nx2 is continuous on R, and the limit function<br />
(1+nx 2 ) 2<br />
g is continuous on R − {0} , then by <strong>The</strong>orem 9.2, the interval I that we<br />
consider does not contains 0. Claim that f n ′ → g uniformly on such interval<br />
I = [a, b] which does not contain 0 as follows.<br />
Consider ∣ ∣ ∣∣∣ 1 − nx 2 ∣∣∣ 1<br />
(1 + nx 2 ) 2 ≤<br />
1 + nx ≤ 1<br />
2 na , 2<br />
so we know that f ′ n → g uniformly on such interval I = [a, b] which does not<br />
contain 0.<br />
9.15 Let f n (x) = (1/n) e −n2 x 2 if x ∈ R, n = 1, 2, ... Prove that f n → 0<br />
uniformly on R, that f ′ n → 0 pointwise on R, but that the convergence of<br />
{f ′ n} is not uniform on any interval containing the origin.<br />
Proof: It is clear that f n → 0 uniformly on R, that f n ′ → 0 pointwise<br />
on R. Assume that f n ′ → 0 uniformly on [a, b] that contains 0. We will prove<br />
that it is impossible as follows.<br />
We may assume that 0 ∈ (a, b) since other cases are similar. Given ε = 1,<br />
e<br />
then there exists a positive integer N ′ such that as n ≥ max ( )<br />
N ′ , 1 b := N<br />
(⇒ 1 ≤ b), we have N<br />
|f n ′ (x) − 0| < 1 for all x ∈ [a, b]<br />
e<br />
which implies that<br />
∣ ∣∣∣<br />
2 Nx<br />
e (Nx)2 ∣ ∣∣∣<br />
< 1 e<br />
for all x ∈ [a, b]<br />
which implies that, let x = 1 N , 2<br />
e < 1 e<br />
which is absurb. So, the convergence of {f ′ n} is not uniform on any interval<br />
containing the origin.<br />
14
9.16 Let {f n } be a sequence of real-valued continuous functions defined<br />
on [0, 1] and assume that f n → f uniformly on [0, 1] . Prove or disprove<br />
∫ 1−1/n<br />
lim<br />
n→∞<br />
0<br />
f n (x) dx =<br />
∫ 1<br />
0<br />
f (x) dx.<br />
Proof: By <strong>The</strong>orem 9.8, we have<br />
∫ 1<br />
lim<br />
n→∞<br />
0<br />
f n (x) dx =<br />
∫ 1<br />
0<br />
f (x) dx. (*)<br />
Note that {f n } is uniform bound, say |f n (x)| ≤ M for all x ∈ [0, 1] and all<br />
n by Exercise 9.1. Hence,<br />
∫ 1<br />
∣ f n (x) dx<br />
∣ ≤ M → 0. (**)<br />
n<br />
1−1/n<br />
Hence, by (*) and (**), we have<br />
∫ 1−1/n<br />
lim<br />
n→∞<br />
0<br />
f n (x) dx =<br />
∫ 1<br />
0<br />
f (x) dx.<br />
9.17 Mathematicinas from Slobbovia decided that the Riemann integral<br />
was too complicated so that they replaced it by Slobbovian integral, defined<br />
as follows: If f is a function defined on the set Q of rational numbers<br />
in [0, 1] , the Slobbovian integral of f, denoted by S (f) , is defined to be the<br />
limit<br />
1<br />
S (f) = lim<br />
n→∞ n<br />
n∑<br />
f<br />
k=1<br />
( k<br />
n)<br />
,<br />
whenever the limit exists. Let {f n } be a sequence of functions such that<br />
S (f n ) exists for each n and such that f n → f uniformly on Q. Prove that<br />
{S (f n )} converges, that S (f) exists, and S (f n ) → S (f) as n → ∞.<br />
Proof: f n → f uniformly on Q, then given ε > 0, there exists a positive<br />
integer N such that as n > m ≥ N, we have<br />
|f n (x) − f (x)| < ε/3 (1)<br />
15
and<br />
So, if n > m ≥ N,<br />
|S (f n ) − S (f m )| =<br />
∣ lim 1<br />
k∑<br />
k→∞ k<br />
j=1<br />
1<br />
k∑<br />
= lim<br />
k→∞ k ∣<br />
|f n (x) − f m (x)| < ε/2. (2)<br />
1<br />
≤ lim<br />
k→∞ k<br />
= ε/2<br />
< ε<br />
j=1<br />
( ( ) ( )) ∣ j j ∣∣∣<br />
f n − f m<br />
k k<br />
( ( ) ( )) ∣ j j ∣∣∣<br />
f n − f m<br />
k k<br />
k∑<br />
ε/2 by (2)<br />
j=1<br />
which implies that {S (f n )} converges since it is a Cauchy sequence. Say its<br />
limit S.<br />
Consider, by (1) as n ≥ N,<br />
1<br />
k<br />
k∑<br />
j=1<br />
[ ( ) ] j<br />
f n − ε/3 ≤ 1 k<br />
k<br />
k∑<br />
( ) j<br />
f ≤ 1 k k<br />
j=1<br />
k∑<br />
j=1<br />
[ ( ) ] j<br />
f n + ε/3<br />
k<br />
which implies that<br />
[<br />
1<br />
k∑<br />
( ) ] j<br />
f n − ε/3 ≤ 1 k k<br />
k<br />
j=1<br />
k∑<br />
( ) j<br />
f ≤<br />
k<br />
j=1<br />
[<br />
1<br />
k<br />
k∑<br />
( ) ] j<br />
f n + ε/3<br />
k<br />
j=1<br />
which implies that, let k → ∞<br />
S (f n ) − ε/3 ≤ lim<br />
k→∞<br />
sup 1 k<br />
k∑<br />
( ) j<br />
f ≤ S (f n ) + ε/3 (3)<br />
k<br />
j=1<br />
and<br />
S (f n ) − ε/3 ≤ lim<br />
k→∞<br />
inf 1 k<br />
k∑<br />
( ) j<br />
f ≤ S (f n ) + ε/3 (4)<br />
k<br />
j=1<br />
16
which implies that<br />
∣ lim sup 1 k∑<br />
( ) j<br />
f − lim inf 1 k∑<br />
( ) ∣ j ∣∣∣<br />
f<br />
k→∞ k k k→∞ k k<br />
j=1<br />
j=1<br />
≤<br />
∣ lim sup 1 k∑<br />
( )<br />
∣<br />
j ∣∣∣∣ f − S (f n )<br />
k→∞ k k ∣ + lim inf 1<br />
k→∞ k<br />
j=1<br />
k∑<br />
( )<br />
j f − S (f n )<br />
k ∣<br />
≤ 2ε by (3) and (4)<br />
3<br />
< ε. (5)<br />
Note that (3)-(5) imply that the existence of S (f) . Also, (3) or (4) implies<br />
that S (f) = S. So, we complete the proof.<br />
9.18 Let f n (x) = 1/ (1 + n 2 x 2 ) if 0 ≤ x ≤ 1, n = 1, 2, ... Prove that {f n }<br />
converges pointwise but not uniformly on [0, 1] . Is term-by term integration<br />
permissible<br />
Proof: It is clear that<br />
lim f n (x) = 0<br />
n→∞<br />
for all x ∈ [0, 1] . If {f n } converges uniformly on [0, 1] , then given ε = 1/3,<br />
there exists a positive integer N such that as n ≥ N, we have<br />
which implies that<br />
|f n (x)| < 1/3 for all x ∈ [0, 1]<br />
∣ ∣∣∣<br />
f N<br />
( 1<br />
N<br />
)∣ ∣∣∣<br />
= 1 2 < 1 3<br />
j=1<br />
which is impossible. So, {f n } converges pointwise but not uniformly on [0, 1] .<br />
Since {f n (x)} is clearly uniformly bounded on [0, 1] , i.e., |f n (x)| ≤ 1<br />
for all x ∈ [0, 1] and n. Hence, by Arzela’s <strong>The</strong>orem, we know that the<br />
sequence of functions can be integrated term by term.<br />
9.19 Prove that ∑ ∞<br />
n=1 x/nα (1 + nx 2 ) converges uniformly on every finite<br />
interval in R if α > 1/2. Is the convergence uniform on R<br />
Proof: By A.P. ≥ G.P., we have<br />
x<br />
∣n α (1 + nx 2 ) ∣ ≤ 1<br />
17<br />
2n α+ 1 2<br />
for all x.
So, by Weierstrass M-test, we have proved that ∑ ∞<br />
n=1 x/nα (1 + nx 2 ) converges<br />
uniformly on R if α > 1/2. Hence, ∑ ∞<br />
n=1 x/nα (1 + nx 2 ) converges<br />
uniformly on every finite interval in R if α > 1/2.<br />
9.20 Prove that the series ∑ ∞<br />
n=1 ((−1)n / √ n) sin (1 + (x/n)) converges<br />
uniformly on every compact subset of R.<br />
Proof: It suffices to show that the series ∑ ∞<br />
n=1 ((−1)n / √ n) sin (1 + (x/n))<br />
converges uniformly on [0, a] . Choose n large enough so that a/n ≤ 1/2, and<br />
therefore sin ( 1 + ( )) (<br />
x<br />
n+1 ≤ sin 1 +<br />
x<br />
n)<br />
for all x ∈ [0, a] . So, if we let fn (x) =<br />
(−1) n / √ n and g n (x) = sin ( 1 + n) x , then by Abel’s test for uniform convergence,<br />
we have proved that the series ∑ ∞<br />
n=1 ((−1)n / √ n) sin (1 + (x/n))<br />
converges uniformly on [0, a] .<br />
Remark: In the proof, we metion something to make the reader get<br />
more. (1) since a compact set K is a bounded set, say K ⊆ [−a, a] , if we can<br />
show the series converges uniformly on [−a, a] , then we have proved it. (2)<br />
<strong>The</strong> interval that we consider is [0, a] since [−a, 0] is similar. (3) Abel’s test<br />
for uniform convergence holds for n ≥ N, where N is a fixed positive<br />
integer.<br />
9.21 Prove that the series ∑ ∞<br />
n=0 (x2n+1 / (2n + 1) − x n+1 / (2n + 2)) converges<br />
pointwise but not uniformly on [0, 1] .<br />
Proof: We show that the series converges pointwise on [0, 1] by considering<br />
two cases: (1) x ∈ [0, 1) and (2) x = 1. Hence, it is trivial. Define<br />
f (x) = ∑ ∞<br />
n=0 (x2n+1 / (2n + 1) − x n+1 / (2n + 2)) , if the series converges<br />
uniformly on [0, 1] , then by <strong>The</strong>orem 9.2, f (x) is continuous on [0, 1] .<br />
However,<br />
f (x) =<br />
{ 1<br />
2<br />
log (1 + x) if x ∈ [0, 1)<br />
log 2 if x = 1<br />
Hence, the series converges not uniformly on [0, 1] .<br />
Remark: <strong>The</strong> function f (x) is found by the following. Given x ∈ [0, 1),<br />
then both<br />
∞∑<br />
t 2n = 1<br />
1 − t and 1 ∞∑<br />
t n 1<br />
=<br />
2 2 2 (1 − t)<br />
n=0<br />
converges uniformly on [0, x] by <strong>The</strong>orem 9.14. So, by <strong>The</strong>orem 9.8, we<br />
n=0<br />
.<br />
18
have<br />
∫ x<br />
0<br />
∞∑<br />
t 2n − 1 2<br />
n=0<br />
∞∑<br />
t n =<br />
n=0<br />
=<br />
∫ x<br />
0<br />
∫ x<br />
0<br />
1<br />
1 − t − 1<br />
2 2 (1 − t) dt<br />
(<br />
1 1<br />
2 1 − t + 1 )<br />
1 + t<br />
= 1 log (1 + x) .<br />
2<br />
− 1 2<br />
( ) 1<br />
dt<br />
1 − t<br />
<strong>And</strong> as x = 1,<br />
∞∑ (<br />
x 2n+1 / (2n + 1) − x n+1 / (2n + 2) )<br />
n=0<br />
=<br />
=<br />
∞∑<br />
n=0<br />
∞∑<br />
n=0<br />
1<br />
2n + 1 − 1<br />
2n<br />
(−1) n+1<br />
n + 1<br />
by <strong>The</strong>orem8.14.<br />
= log 2 by Abel’s Limit <strong>The</strong>orem.<br />
9.22 Prove that ∑ a n sin nx and ∑ b n cos nx are uniformly convergent on<br />
R if ∑ |a n | converges.<br />
Proof: It is trivial by Weierstrass M-test.<br />
9.23 Let {a n } be a decreasing sequence of positive terms. Prove that<br />
the series ∑ a n sin nx converges uniformly on R if, and only if, na n → 0 as<br />
n → ∞.<br />
Proof: (⇒) Suppose that the series ∑ a n sin nx converges uniformly on<br />
R, then given ε > 0, there exists a positive integer N such that as n ≥ N,<br />
we have<br />
∣ ∣∣∣∣ 2n−1<br />
∑<br />
a k sin kx<br />
< ε. (*)<br />
∣<br />
Choose x = 1<br />
2n , then sin 1 2<br />
k=n<br />
≤ sin kx ≤ sin 1. Hence, as n ≥ N, we always<br />
19
have, by (*)<br />
2n−1<br />
∑<br />
2n−1<br />
(ε >)<br />
a k sin kx<br />
∣<br />
∣ = ∑<br />
a k sin kx<br />
k=n<br />
≥<br />
=<br />
k=n<br />
2n−1<br />
∑<br />
a 2n sin 1 2 since a k > 0 and a k ↘<br />
k=n<br />
( 1<br />
2 sin 1 2)<br />
(2na 2n ) .<br />
That is, we have proved that 2na 2n → 0 as n → ∞. Similarly, we also have<br />
(2n − 1) a 2n−1 → 0 as n → ∞. So, we have proved that na n → 0 as n → ∞.<br />
(⇐) Suppose that na n → 0 as n → ∞, then given ε > 0, there exists a<br />
positive integer n 0 such that as n ≥ n 0 , we have<br />
ε<br />
|na n | = na n <<br />
2 (π + 1) . (*)<br />
In order to show the uniform convergence of ∑ ∞<br />
n=1 a n sin nx on R, it suffices<br />
to show the uniform convergence of ∑ ∞<br />
n=1 a n sin nx on [0, π] . So, if we can<br />
show that as n ≥ n 0<br />
n+p<br />
∑<br />
a k sin kx<br />
< ε for all x ∈ [0, π] , and all p ∈ N<br />
∣<br />
∣<br />
k=n+1<br />
then we complete [ ] it. We consider two cases as follows. (n ≥ n 0 )<br />
π<br />
As x ∈ 0, , then<br />
n+p<br />
∣<br />
∣<br />
∣<br />
n+p<br />
∑<br />
k=n+1<br />
n+p<br />
a k sin kx<br />
∣ = ∑<br />
≤<br />
=<br />
k=n+1<br />
n+p<br />
∑<br />
k=n+1<br />
n+p<br />
∑<br />
k=n+1<br />
a k sin kx<br />
a k kx by sin kx ≤ kx if x ≥ 0<br />
(ka k ) x<br />
ε pπ<br />
≤<br />
by (*)<br />
2 (π + 1) n + p<br />
< ε.<br />
20
[ ]<br />
π<br />
<strong>And</strong> as x ∈ , π , then<br />
n+p<br />
∣<br />
n+p<br />
∑<br />
k=n+1<br />
a k sin kx<br />
∣ ≤<br />
≤<br />
≤<br />
≤<br />
≤<br />
m<br />
∑<br />
k=n+1<br />
a k sin kx +<br />
∣<br />
n+p<br />
∑<br />
k=m+1<br />
[ π<br />
]<br />
a k sin kx<br />
∣ , where m = x<br />
m∑<br />
a k kx + 2a m+1<br />
sin x by Summation by parts<br />
k=n+1<br />
2<br />
ε<br />
2 (π + 1) (m − n) x + 2a m+1<br />
sin x 2<br />
ε<br />
π<br />
2 (π + 1) mx + 2a m+1<br />
x by 2x [0,<br />
π ≤ sin x if x ∈ π 2<br />
ε<br />
2 (π + 1) π + 2a m+1 (m + 1)<br />
< ε 2 + 2 ε<br />
2 (π + 1)<br />
< ε.<br />
Hence, ∑ ∞<br />
n=1 a n sin nx converges uniformly on R.<br />
Remark: (1) In the proof (⇐), if we can make sure that na n ↘ 0, then<br />
we can use the supplement on the convergnce of series in Ch8, (C)-<br />
(6) to show the uniform convergence of ∑ ∞<br />
n=1 a n sin nx = ∑ ∞<br />
n=1 (na n) ( )<br />
sin nx<br />
n<br />
by Dirichlet’s test for uniform convergence.<br />
(2)<strong>The</strong>re are similar results; we write it as references.<br />
(a) Suppose a n ↘ 0, then for each α ∈ ( 0, π 2<br />
)<br />
,<br />
∑ ∞<br />
n=1 a n cos nx and<br />
∑ ∞<br />
n=1 a n sin nx converges uniformly on [α, 2π − α] .<br />
Proof: <strong>The</strong> proof follows from (12) and (13) in <strong>The</strong>orem 8.30 and<br />
Dirichlet’s test for uniform convergence. So, we omit it. <strong>The</strong> reader<br />
can see the textbook, example in pp 231.<br />
(b) Let {a n } be a decreasing sequence of positive terms. ∑ ∞<br />
n=1 a n cos nx<br />
uniformly converges on R if and only if ∑ ∞<br />
n=1 a n converges.<br />
Proof: (⇒) Suppose that ∑ ∞<br />
n=1 a n cos nx uniformly converges on R, then<br />
let x = 0, then we have ∑ ∞<br />
n=1 a n converges.<br />
(⇐) Suppose that ∑ ∞<br />
n=1 a n converges, then by Weierstrass M-test, we<br />
have proved that ∑ ∞<br />
n=1 a n cos nx uniformly converges on R.<br />
]<br />
21
9.24 Given a convergent series ∑ ∞<br />
∑ n=1 a n. Prove that the Dirichlet series<br />
∞<br />
n=1 a nn −s converges uniformly<br />
∑<br />
on the half-infinite interval 0 ≤ s < +∞.<br />
Use this to prove that lim ∞<br />
s→0 +<br />
n=1 a nn −s = ∑ ∞<br />
n=1 a n.<br />
Proof: Let f n (s) = ∑ n<br />
k=1 a k and g n (s) = n −s , then by Abel’s test for<br />
uniform convergence, we have proved that the Dirichlet series ∑ ∞<br />
n=1 a nn −s<br />
converges uniformly on the half-infinite<br />
∑<br />
interval 0 ≤ s < +∞. <strong>The</strong>n by<br />
<strong>The</strong>orem 9.2, we know that lim ∞<br />
s→0 +<br />
n=1 a nn −s = ∑ ∞<br />
n=1 a n.<br />
9.25 Prove that the series ζ (s) = ∑ ∞<br />
n=1 n−s converges uniformly on every<br />
half-infinite interval 1 + h ≤ s < +∞, where h > 0. Show that the equation<br />
ζ ′ (s) = −<br />
∞∑<br />
n=1<br />
log n<br />
n s<br />
is valid for each s > 1 and obtain a similar formula for the kth derivative<br />
ζ (k) (s) .<br />
Proof: Since n −s ≤ n −(1+h) for all s ∈ [1 + h, ∞), we know that ζ (s) =<br />
∑ ∞<br />
n=1 n−s converges uniformly on every half-infinite interval 1+h ≤ s < +∞<br />
by Weierstrass M-test. Define T n (s) = ∑ n<br />
k=1 k−s , then it is clear that<br />
<strong>And</strong><br />
1. For each n, T n (s) is differentiable on [1 + h, ∞),<br />
3. T ′ n (s) = −<br />
n∑<br />
k=1<br />
2. lim<br />
n→∞<br />
T n (2) = π2<br />
6 .<br />
log k<br />
k s converges uniformly on [1 + h, ∞)<br />
by Weierstrass M-test. Hence, we have proved that<br />
ζ ′ (s) = −<br />
∞∑<br />
n=1<br />
log n<br />
n s<br />
by <strong>The</strong>orem 9.13. By Mathematical Induction, we know that<br />
ζ (k) (s) = (−1) k<br />
∞<br />
∑<br />
n=1<br />
(log n) k<br />
n s .<br />
22
0.1 Supplement on some results on Weierstrass M-<br />
test.<br />
1. In the textbook, pp 224-223, there is a surprising result called Spacefilling<br />
curve. In addition, note the proof is related with Cantor set in<br />
exercise 7. 32 in the textbook.<br />
2. <strong>The</strong>re exists a continuous function defined on R which is nowhere<br />
differentiable. <strong>The</strong> reader can see the book, Principles of Mathematical<br />
Analysis by Walter Rudin, pp 154.<br />
Remark: <strong>The</strong> first example comes from Bolzano in 1834, however, he<br />
did NOT give a proof. In fact, he only found the function f : D → R that<br />
he constructed is not differentiable on D ′ (⊆ D) where D ′ is countable and<br />
dense in D. Although the function f is the example, but he did not find the<br />
fact.<br />
In 1861, Riemann gave<br />
g (x) =<br />
∞∑<br />
n=1<br />
sin (n 2 πx)<br />
n 2<br />
as an example. However, Reimann did NOT give a proof in his life until<br />
1916, the proof is given by G. Hardy.<br />
In 1860, Weierstrass gave<br />
h (x) =<br />
∞∑<br />
a n cos (b n πx) , b is odd, 0 < a < 1, and ab > 1 + 3π 2 ,<br />
n=1<br />
until 1875, he gave the proof. <strong>The</strong> fact surprises the world of Math, and<br />
produces many examples. <strong>The</strong>re are many researches related with it until<br />
now 2003.<br />
Mean Convergence<br />
9.26 Let f n (x) = n 3/2 xe −n2 x 2 . Prove that {f n } converges pointwise to 0<br />
on [−1, 1] but that l.i.m. n→∞ f n ≠ 0 on [−1, 1] .<br />
Proof: It is clear that {f n } converges pointwise to 0 on [−1, 1] , so it<br />
23
emains to show that l.i.m. n→∞ f n ≠ 0 on [−1, 1] . Consider<br />
∫ 1<br />
−1<br />
f 2 n (x) dx = 2<br />
∫ 1<br />
0<br />
n 3 x 2 e −2n2 x 2 dx since f 2 n (x) is an even function on [−1, 1]<br />
= √ 1 ∫ √ 2n<br />
y 2 e −y2 dy by Change of Variable, let y = √ 2nx<br />
2<br />
= 1<br />
−2 √ 2<br />
0<br />
∫ √ 2n<br />
0<br />
yd<br />
(e −y2)<br />
[<br />
= 1 ∣√ ∣∣ 2n<br />
∫ √ ]<br />
2n<br />
−2 √ ye −y2 − e −y2 dy<br />
2<br />
0 0<br />
√ ∫ π<br />
∞<br />
√ π<br />
→<br />
4 √ 2 since e −x2 dx = by Exercise 7. 19.<br />
2<br />
0<br />
So, l.i.m. n→∞ f n ≠ 0 on [−1, 1] .<br />
9.27 Assume that {f n } converges pointwise to f on [a, b] and that<br />
l.i.m. n→∞ f n = g on [a, b] . Prove that f = g if both f and g are continuous<br />
on [a, b] .<br />
Proof: Since l.i.m. n→∞ f n = g on [a, b] , given ε k = 1<br />
2 k , there exists a n k<br />
such that<br />
Define<br />
then<br />
∫ b<br />
a<br />
|f nk (x) − g (x)| p dx ≤ 1 2 k , where p > 0<br />
h m (x) =<br />
m∑<br />
k=1<br />
∫ x<br />
a. h m (x) ↗ as x ↗<br />
b. h m (x) ≤ h m+1 (x)<br />
a<br />
|f nk (t) − g (t)| p dt,<br />
c. h m (x) ≤ 1 for all m and all x.<br />
So, we obtain h m (x) → h (x) as m → ∞, h (x) ↗ as x ↗, and<br />
h (x) − h m (x) =<br />
∞∑<br />
k=m+1<br />
∫ x<br />
a<br />
|f nk (t) − g (t)| p dt ↗ as x ↗<br />
24
which implies that<br />
h (x + t) − h (x)<br />
t<br />
≥ h m (x + t) − h m (x)<br />
t<br />
for all m. (*)<br />
Since h and h m are increasing, we have h ′ and h ′ m exists a.e. on [a, b] . Hence,<br />
by (*)<br />
m∑<br />
h ′ m (x) = |f nk (t) − g (t)| p ≤ h ′ (x) a.e. on [a, b]<br />
k=1<br />
which implies that<br />
∞∑<br />
|f nk (t) − g (t)| p exists a.e. on [a, b] .<br />
k=1<br />
So, f nk (t) → g (t) a.e. on [a, b] . In addition, f n → f on [a, b] . <strong>The</strong>n we<br />
conclude that f = g a.e. on [a, b] . Since f and g are continuous on [a, b] , we<br />
have<br />
∫ b<br />
a<br />
|f − g| dx = 0<br />
which implies that f = g on [a, b] . In particular, as p = 2, we have f = g.<br />
Remark: (1) A property is said to hold almost everywhere on a set<br />
S (written: a.e. on S) if it holds everywhere on S except for a set of measurer<br />
zero. Also, see the textbook, pp 254.<br />
(2) In this proof, we use the theorem which states: A monotonic function<br />
h defined on [a, b] , then h is differentiable a.e. on [a, b] . <strong>The</strong> reader can<br />
see the book, <strong>The</strong> reader can see the book, Measure and Integral (An<br />
Introduction to <strong>Real</strong> Analysis) written by Richard L. Wheeden and<br />
Antoni Zygmund, pp 113.<br />
(3) <strong>The</strong>re is another proof by using Fatou’s lemma: Let {f k } be a<br />
measruable function defined on a measure set E. If f k ≥ φ a.e. on E and<br />
φ ∈ L (E) , then ∫<br />
∫<br />
lim k ≤ lim inf f k .<br />
E<br />
k→∞ k→∞<br />
E<br />
Proof: It suffices to show that f nk (t) → g (t) a.e. on [a, b] . Since<br />
l.i.m. n→∞ f n = g on [a, b] , and given ε > 0, there exists a n k such that<br />
∫ b<br />
a<br />
|f nk − g| 2 dx < 1 2 k<br />
25
which implies that<br />
∫ b m∑<br />
a<br />
k=1<br />
|f nk − g| 2 dx <<br />
which implies that, by Fatou’s lemma,<br />
m∑<br />
k=1<br />
1<br />
2 k<br />
∫ b<br />
a<br />
lim inf ∑ m ∫ b m∑<br />
|f nk − g| 2 dx ≤ lim inf |f nk − g| 2 dx<br />
m→∞ m→∞<br />
k=1<br />
a<br />
k=1<br />
∞∑<br />
∫ b<br />
= |f nk − g| 2 dx < 1.<br />
k=1<br />
a<br />
That is,<br />
which implies that<br />
∫ b ∞∑<br />
a<br />
k=1<br />
|f nk − g| 2 dx < 1<br />
∞∑<br />
|f nk − g| 2 < ∞ a.e. on [a, b]<br />
k=1<br />
which implies that f nk → g a.e. on [a, b] .<br />
Note: <strong>The</strong> reader can see the book, Measure and Integral (An Introduction<br />
to <strong>Real</strong> Analysis) written by Richard L. Wheeden and<br />
Antoni Zygmund, pp 75.<br />
(4) <strong>The</strong>re is another proof by using Egorov’s <strong>The</strong>orem: Let {f k } be a<br />
measurable functions defined on a finite measurable set E with finite limit<br />
function f. <strong>The</strong>n given ε > 0, there exists a closed set F (⊆ E) , where<br />
|E − F | < ε such that<br />
f k → f uniformly on F.<br />
Proof: If f ≠ g on [a, b] , then h := |f − g| ≠ 0 on [a, b] . By continuity<br />
of h, there exists a compact subinterval [c, d] such that |f − g| ≠ 0. So, there<br />
exists m > 0 such that h = |f − g| ≥ m > 0 on [c, d] . Since<br />
∫ b<br />
a<br />
|f n − g| 2 dx → 0 as n → ∞,<br />
26
we have<br />
∫ d<br />
c<br />
|f n − g| 2 dx → 0 as n → ∞.<br />
then by Egorov’s <strong>The</strong>orem, given ε > 0, there exists a closed susbet F of<br />
[c, d] , where |[c, d] − F | < ε such that<br />
which implies that<br />
f n → f uniformly on F<br />
0 = lim |f n − g|<br />
n→∞<br />
∫F<br />
2 dx<br />
∫<br />
= lim |f n − g| 2 dx<br />
F<br />
n→∞<br />
∫<br />
= |f − g| 2 dx ≥ m 2 |F |<br />
F<br />
which implies that |F | = 0. If we choose ε < d−c, then we get a contradiction.<br />
<strong>The</strong>refore, f = g on [a, b] .<br />
Note: <strong>The</strong> reader can see the book, Measure and Integral (An Introduction<br />
to <strong>Real</strong> Analysis) written by Richard L. Wheeden and<br />
Antoni Zygmund, pp 57.<br />
9.28 Let f n (x) = cos n x if 0 ≤ x ≤ π.<br />
(a) Prove that l.i.m. n→∞ f n<br />
converge.<br />
= 0 on [0, π] but that {f n (π)} does not<br />
Proof: It is clear that {f n (π)} does not converge since f n (π) = (−1) n .<br />
It remains to show that l.i.m. n→∞ f n = 0 on [0, π] . Consider cos 2n x := g n (x)<br />
on [0, π] , then it is clear that {g n (x)} is boundedly convergent with limit<br />
function<br />
{ 0 if x ∈ (0, π)<br />
g =<br />
1 if x = 0 or π .<br />
Hence, by Arzela’s <strong>The</strong>orem,<br />
∫ π<br />
lim<br />
n→∞<br />
0<br />
So, l.i.m. n→∞ f n = 0 on [0, π] .<br />
cos 2n xdx =<br />
27<br />
∫ π<br />
0<br />
g (x) dx = 0.
(b) Prove that {f n } converges pointwise but not uniformly on [0, π/2] .<br />
Proof: Note that each f n (x) is continuous on [0, π/2] , and the limit<br />
function<br />
{ 0 if x ∈ (0, π/2]<br />
f =<br />
.<br />
1 if x = 0<br />
Hence, by <strong>The</strong>orem9.2, we know that {f n } converges pointwise but not<br />
uniformly on [0, π/2] .<br />
9.29 Let f n (x) = 0 if 0 ≤ x ≤ 1/n or 2/n ≤ x ≤ 1, and let f n (x) = n if<br />
1/n < x < 2/n. Prove that {f n } converges pointwise to 0 on [0, 1] but that<br />
l.i.m. n→∞ f n ≠ 0 on [0, 1] .<br />
Proof: It is clear that {f n } converges pointwise to 0 on [0, 1] . In order<br />
to show that l.i.m. n→∞ f n ≠ 0 on [0, 1] , it suffices to note that<br />
∫ 1<br />
Hence, l.i.m. n→∞ f n ≠ 0 on [0, 1] .<br />
Power series<br />
0<br />
f n (x) dx = 1 for all n.<br />
9.30 If r is the radius of convergence if ∑ a n (z − z 0 ) n , where each a n ≠ 0,<br />
show that<br />
∣ ∣ ∣ ∣∣∣<br />
lim inf a n ∣∣∣ ≤ r ≤ lim sup<br />
a n ∣∣∣<br />
n→∞ a n+1 n→∞ ∣ .<br />
a n+1<br />
Proof: By Exercise 8.4, we have<br />
1<br />
∣ lim n→∞ sup ∣ a n+1 ∣∣<br />
≤ r =<br />
a n<br />
1<br />
lim n→∞ sup |a n | 1 n<br />
≤<br />
1<br />
lim n→∞ inf<br />
∣ ∣ a n+1 ∣∣<br />
.<br />
a n<br />
Since<br />
and<br />
1<br />
∣<br />
lim n→∞ sup<br />
1<br />
lim n→∞ inf<br />
∣<br />
∣ a n+1<br />
a n<br />
∣ a n+1<br />
a n<br />
∣ ∣∣<br />
= lim<br />
n→∞<br />
inf<br />
∣ ∣∣<br />
= lim<br />
n→∞<br />
sup<br />
∣ a n ∣∣∣<br />
∣a n+1<br />
∣ a n ∣∣∣<br />
∣ ,<br />
a n+1<br />
28
we complete it.<br />
9.31 Given that two power series ∑ a n z n has radius of convergence 2.<br />
Find the radius convergence of each of the following series: In (a) and (b), k<br />
is a fixed positive integer.<br />
(a) ∑ ∞<br />
n=0 ak nz n<br />
Proof: Since<br />
2 =<br />
we know that the radius of ∑ ∞<br />
n=0 ak nz n is<br />
1<br />
, (*)<br />
1/n<br />
lim n→∞ sup |a n |<br />
1<br />
lim n→∞ sup |a k n| 1/n = 1<br />
(lim n→∞ sup |a n | 1/n) k = 2k .<br />
(b) ∑ ∞<br />
n=0 a nz kn<br />
Proof: Consider<br />
which implies that<br />
lim sup ∣ an z kn∣ ∣ 1/n = lim sup |a n | 1/n |z| k < 1<br />
n→∞ n→∞<br />
(<br />
) 1/k<br />
1<br />
|z| <<br />
= 2 1/k by (*).<br />
lim n→∞ sup |a n | 1/n<br />
So, the radius of ∑ ∞<br />
n=0 a nz kn is 2 1/k .<br />
(c) ∑ ∞<br />
n=0 a nz n2<br />
Proof: Consider<br />
∣<br />
∣ ∣∣<br />
1/n<br />
lim sup ∣a n z n2 = lim sup |a n | 1/n |z| n<br />
n→∞<br />
and claim that the radius of ∑ ∞<br />
n=0 a nz n2 is 1 as follows.<br />
If |z| < 1, it is clearly seen that the series converges. However, if |z| > 1,<br />
lim sup |a n| 1/n lim inf |z| n ≤ lim sup |a n | 1/n |z| n<br />
n→∞ n→∞ n→∞<br />
29
which impliest that<br />
lim sup |a n| 1/n |z| n = +∞.<br />
n→∞<br />
so, the series diverges. From above, we have proved the claim.<br />
9.32 Given a power series ∑ a n x n whose coefficents are related by an<br />
equation of the form<br />
a n + Aa n−1 + Ba n−2 = 0 (n = 2, 3, ...).<br />
Show that for any x for which the series converges, its sum is<br />
a 0 + (a 1 + Aa 0 ) x<br />
1 + Ax + Bx 2 .<br />
Proof: Consider<br />
which implies that<br />
n=2<br />
which implies that<br />
n=0<br />
∞∑<br />
(a n + Aa n−1 + Ba n−2 ) x n = 0<br />
n=2<br />
∞∑<br />
∞∑<br />
a n x n + Ax a n−1 x n−1 + Bx 2<br />
n=2<br />
∞∑<br />
∞∑<br />
a n x n + Ax a n x n + Bx 2<br />
which implies that<br />
n=0<br />
∞∑<br />
n=0<br />
∞<br />
∑<br />
n=0<br />
∞<br />
∑<br />
n=2<br />
a n x n = a 0 + (a 1 + Aa 0 ) x<br />
1 + Ax + Bx 2 .<br />
a n−2 x n−2 = 0<br />
a n x n = a 0 + a 1 x + Aa 0 x<br />
Remark: We prove that for any x for which the series converges, then<br />
1 + Ax + Bx 2 ≠ 0 as follows.<br />
Proof: Consider<br />
( ) ∑<br />
∞<br />
1 + Ax + Bx<br />
2<br />
a n x n = a 0 + (a 1 + Aa 0 ) x,<br />
n=0<br />
30
if x = λ (≠ 0) is a root of 1 + Ax + Bx 2 , and ∑ ∞<br />
n=0 a nλ n exists, we have<br />
1 + Aλ + Bλ 2 = 0 and a 0 + (a 1 + Aa 0 ) λ = 0.<br />
Note that a 1 + Aa 0 ≠ 0, otherwise, a 0 = 0 (⇒ a 1 = 0) , and therefore, a n =<br />
0 for all n. <strong>The</strong>n there is nothing to prove it. So, put λ = −a 0<br />
a 1 +Aa 0<br />
into<br />
1 + Aλ + Bλ 2 = 0, we then have<br />
a 2 1 = a 0 a 2 .<br />
Note that a 0 ≠ 0, otherwise, a 1 = 0 and a 2 = 0. Similarly, a 1 ≠ 0, otherwise,<br />
we will obtain a trivial thing. Hence, we may assume that all a n ≠ 0 for all<br />
n. So,<br />
a 2 2 = a 1 a 3 .<br />
<strong>And</strong> it is easy to check that a n = a 0<br />
1<br />
λ n for all n ≥ N. <strong>The</strong>refore, ∑ a n λ n =<br />
∑<br />
a0 diverges. So, for any x for whcih the series converges, we have 1+Ax+<br />
Bx 2 ≠ 0.<br />
9.33 Let f (x) = e −1/x2 if x ≠ 0, f (0) = 0.<br />
(a) Show that f (n) (0) exists for all n ≥ 1.<br />
Proof: By Exercise 5.4, we complete it.<br />
(b) Show that the Taylor’s series about 0 generated by f converges everywhere<br />
on R but that it represents f only at the origin.<br />
Proof: <strong>The</strong> Taylor’s series about 0 generated by f is<br />
∞∑<br />
n=0<br />
f (n) (0)<br />
x n =<br />
n!<br />
∞∑<br />
0x n = 0.<br />
So, it converges everywhere on R but that it represents f only at the origin.<br />
Remark: It is an important example to tell us that even for functions<br />
f ∈ C ∞ (R) , the Taylor’s series about c generated by f may NOT represent<br />
f on some open interval. Also see the textbook, pp 241.<br />
9.34 Show that the binomial series (1 + x) α = ∑ ∞<br />
n=0 (α n) x n exhibits the<br />
following behavior at the points x = ±1.<br />
(a) If x = −1, the series converges for α ≥ 0 and diverges for α < 0.<br />
n=0<br />
31
Proof: If x = −1, we consider three cases: (i) α < 0, (ii) α = 0, and (iii)<br />
α > 0.<br />
(i) As α < 0, then<br />
∞∑<br />
( α n) (−1) n =<br />
n=0<br />
∞∑<br />
n=0<br />
(−1) n α (α − 1) · · · (α − n + 1)<br />
,<br />
n!<br />
say a n = (−1) n α(α−1)···(α−n+1)<br />
n!<br />
, then a n ≥ 0 for all n, and<br />
a n<br />
1/n<br />
=<br />
−α (−α + 1) · · · (−α + n − 1)<br />
(n − 1)!<br />
≥ −α > 0 for all n.<br />
Hence, ∑ ∞<br />
n=0 (α n) (−1) n diverges.<br />
(ii) As α = 0, then the series is clearly convergent.<br />
(iii) As α > 0, define a n = n (−1) n ( α n) , then<br />
a n+1<br />
a n<br />
= n − α<br />
n<br />
≥ 1 if n ≥ [α] + 1. (*)<br />
It means that a n > 0 for all n ≥ [α] + 1 or a n < 0 for all n ≥ [α] + 1. Without<br />
loss of generality, we consider a n > 0 for all n ≥ [α] + 1 as follows.<br />
Note that (*) tells us that<br />
and<br />
So,<br />
m∑<br />
n=[α]+1<br />
a n > a n+1 > 0 ⇒ lim<br />
n→∞<br />
a n exists.<br />
a n − a n+1 = α (−1) n ( α n) .<br />
(−1) n ( α n) = 1 α<br />
m∑<br />
n=[α]+1<br />
(a n − a n+1 ) .<br />
By <strong>The</strong>orem 8.10, we have proved the convergence of the series ∑ ∞<br />
n=0 (α n) (−1) n .<br />
(b) If x = 1, the series diverges for α ≤ −1, converges conditionally for α<br />
in the interval −1 < α < 0, and converges absolutely for α ≥ 0.<br />
Proof: If x = 1, we consider four cases as follows: (i) α ≤ −1, (ii)<br />
−1 < −α < 0, (iii) α = 0, and (iv) α > 0 :<br />
(i) As α ≤ −1, say a n = α(α−1)···(α−n+1)<br />
n!<br />
. <strong>The</strong>n<br />
|a n | =<br />
−α (−α + 1) · · · (−α + n − 1)<br />
n!<br />
32<br />
≥ 1 for all n.
So, the series diverges.<br />
(ii) As −1 < α < 0, say a n = α(α−1)···(α−n+1)<br />
n!<br />
. <strong>The</strong>n a n = (−1) n b n , where<br />
with<br />
b n =<br />
b n+1<br />
b n<br />
−α (−α + 1) · · · (−α + n − 1)<br />
n!<br />
= n − α<br />
n<br />
> 0.<br />
< 1 since − 1 < −α < 0<br />
which implies that {b n } is decreasing with limit L. So, if we can show L = 0,<br />
then ∑ a n converges by <strong>The</strong>orem 8.16.<br />
Rewrite<br />
n∏<br />
(<br />
b n = 1 − α + 1 )<br />
k<br />
k=1<br />
and since ∑ α+1<br />
k<br />
diverges, then by <strong>The</strong>roem 8.55, we have proved L = 0.<br />
In order to show the convergence is conditionally, it suffices to show the<br />
divergence of ∑ b n . <strong>The</strong> fact follows from<br />
b n<br />
1/n<br />
=<br />
−α (−α + 1) · · · (−α + n − 1)<br />
(n − 1)!<br />
≥ −α > 0.<br />
(iii) As α = 0, it is clearly that the series converges absolutely.<br />
(iv) As α > 0, we consider ∑ |( α n)| as follows. Define a n = |( α n)| , then<br />
a n+1<br />
a n<br />
= n − α<br />
n + 1<br />
< 1 if n ≥ [α] + 1.<br />
It implies that na n − (n + 1) a n = αa n and (n + 1) a n+1 < na n . So, by<br />
<strong>The</strong>roem 8.10,<br />
∑<br />
an = 1 ∑<br />
nan − (n + 1) a n<br />
α<br />
converges since lim n→∞ na n exists. So, we have proved that the series converges<br />
absolutely.<br />
9.35 Show that ∑ a n x n converges uniformly on [0, 1] if ∑ a n converges.<br />
Use this fact to give another proof of Abel’s limit theorem.<br />
Proof: Define f n (x) = a n on [0, 1] , then it is clear that ∑ f n (x) converges<br />
uniformly on [0, 1] . In addition, let g n (x) = x n , then g n (x) is unifomrly<br />
bouned with g n+1 (x) ≤ g n (x) . So, by Abel’s test for uniform convergence,<br />
33
∑<br />
an x n converges uniformly on [0, 1] . Now, we give another proof of Abel’s<br />
Limit <strong>The</strong>orem as follows. Note that each term of ∑ a n x n is continuous<br />
on [0, 1] and the convergence is uniformly on [0, 1] , so by <strong>The</strong>orem 9.2, the<br />
power series is continuous on [0, 1] . That is, we have proved Abel’s Limit<br />
<strong>The</strong>orem:<br />
lim<br />
x→1 − ∑<br />
an x n = ∑ a n .<br />
9.36 If each a n > 0 and ∑ a n diverges, show that ∑ a n x n → +∞ as<br />
x → 1 − . (Assume ∑ a n x n converges for |x| < 1.)<br />
Proof: Given M > 0, if we can find a y near 1 from the left such that<br />
∑<br />
an y n ≥ M, then for y ≤ x < 1, we have<br />
M ≤ ∑ a n y n ≤ ∑ a n x n .<br />
That is, lim x→1 −<br />
∑<br />
an x n = +∞.<br />
Since ∑ a n diverges, there is a positive integer p such that<br />
p∑<br />
a k ≥ 2M > M. (*)<br />
k=1<br />
Define f n (x) = ∑ n<br />
k=1 a kx k , then by continuity of each f n , given 0 < ε (< M) ,<br />
there exists a δ n > 0 such that as x ∈ [δ n , 1), we have<br />
n∑<br />
a k − ε <<br />
k=1<br />
n∑<br />
a k x k <<br />
k=1<br />
n∑<br />
a k + ε (**)<br />
k=1<br />
By (*) and (**), we proved that as y = δ p<br />
M ≤<br />
p∑<br />
a k − ε <<br />
k=1<br />
p∑<br />
a k y k .<br />
k=1<br />
Hence, we have proved it.<br />
9.37 If each a n > 0 and if lim x→1 −<br />
∑<br />
an x n exists and equals A, prove<br />
that ∑ a n converges and has the sum A. (Compare with <strong>The</strong>orem 9.33.)<br />
Proof: By Exercise 9.36, we have proved the part, ∑ a n converges. In<br />
order to show ∑ a n = A, we apply Abel’s Limit <strong>The</strong>orem to complete it.<br />
34
9.38 For each real t, define f t (x) = xe xt / (e x − 1) if x ∈ R, x ≠ 0,<br />
f t (0) = 1.<br />
(a) Show that there is a disk B (0; δ) in which f t is represented by a power<br />
series in x.<br />
Proof: First, we note that ex −1<br />
= ∑ ∞ x n<br />
x n=0<br />
:= p (x) , then p (0) = 1 ≠<br />
(n+1)!<br />
0. So, by <strong>The</strong>orem 9. 26, there exists a disk B (0; δ) in which the reciprocal<br />
of p has a power series exapnsion of the form<br />
∞<br />
1<br />
p (x) = ∑<br />
q n x n .<br />
So, as x ∈ B (0; δ) by <strong>The</strong>orem 9.24.<br />
n=0<br />
f t (x) = xe xt / (e x − 1)<br />
( ∞<br />
) (<br />
∑ (xt) n ∞<br />
)<br />
∑ x n<br />
=<br />
n! (n + 1)!<br />
n=0<br />
n=0<br />
∞∑<br />
= r n (t) x n .<br />
n=0<br />
(b) Define P 0 (t) , P 1 (t) , P 2 (t) , ..., by the equation<br />
and use the identity<br />
f t (x) =<br />
n=0<br />
∞∑<br />
n=0<br />
P n (t) xn<br />
, if x ∈ B (0; δ) ,<br />
n!<br />
∞∑<br />
P n (t) xn<br />
n! = ∑ ∞ etx P n (0) xn<br />
n!<br />
to prove that P n (t) = ∑ n<br />
k=0 (n k ) P k (0) t n−k .<br />
Proof: Since<br />
f t (x) =<br />
∞∑<br />
n=0<br />
n=0<br />
P n (t) xn<br />
n! = etx x<br />
e x − 1 ,<br />
35
and<br />
f 0 (x) =<br />
n=0<br />
∞∑<br />
n=0<br />
P n (0) xn<br />
n! = x<br />
e x − 1 .<br />
So, we have the identity<br />
∞∑<br />
P n (t) xn<br />
n! = ∑ ∞ etx P n (0) xn<br />
n! .<br />
n=0<br />
Use the identity with e tx = ∑ ∞ t n<br />
n=0 n! xn , then we obtain<br />
n∑<br />
which implies that<br />
P n (t)<br />
n!<br />
=<br />
k=0<br />
= 1 n!<br />
P n (t) =<br />
t n−k P k (0)<br />
(n − k)! k!<br />
n∑<br />
( n k) P k (0) t n−k<br />
k=0<br />
n∑<br />
( n k) P k (0) t n−k .<br />
k=0<br />
This shows that each function P n is a polynomial. <strong>The</strong>re are the Bernoulli<br />
polynomials. <strong>The</strong> numbers B n = P n (0) (n = 0, 1, 2, ...) are called the<br />
Bernoulli numbers. Derive the following further properties:<br />
(c) B 0 = 1, B 1 = − 1 2 , ∑ n−1<br />
k=0 (n k ) B k = 0, if n = 2, 3, ...<br />
Proof: Since 1 = p(x) , where p (x) := ∑ ∞ x n<br />
p(x) n=0<br />
, and 1 := ∑ ∞<br />
(n+1)! p(x) n=0 P n (0) xn . n!<br />
So,<br />
where<br />
1<br />
1 = p (x)<br />
p (x)<br />
∞∑ x n<br />
=<br />
(n + 1)!<br />
n=0<br />
∞∑<br />
= C n x n<br />
C n =<br />
n=0<br />
1<br />
(n + 1)!<br />
n∑<br />
k=0<br />
36<br />
∞∑<br />
n=0<br />
P n (0) xn<br />
n!<br />
( n+1<br />
)<br />
k Pk (0) .
So,<br />
B 0 = P 0 (0) = C 0 = 1,<br />
B 1 = P 1 (0) = C 1 − P 0 (0)<br />
= − 1 2 2 , by C 1 = 0<br />
and note that C n = 0 for all n ≥ 1, we have<br />
0 = C n−1<br />
= 1 ∑n−1<br />
( n<br />
n!<br />
k) P k (0)<br />
k=0<br />
= 1 ∑n−1<br />
( n<br />
n!<br />
k) B k for all n ≥ 2.<br />
k=0<br />
(d) P ′ n (t) = nP n−1 (t) , if n = 1, 2, ...<br />
Proof: Since<br />
P ′ n (t) =<br />
n∑<br />
( n k) P k (0) (n − k) t n−k−1<br />
k=0<br />
n−1<br />
∑<br />
= ( n k) P k (0) (n − k) t n−k−1<br />
k=0<br />
n−1<br />
∑<br />
=<br />
k=0<br />
n−1<br />
n! (n − k)<br />
k! (n − k)! P k (0) t (n−1)−k<br />
∑ (n − 1)!<br />
= n<br />
k! (n − 1 − k)! P k (0) t (n−1)−k<br />
k=0<br />
n−1<br />
∑<br />
= n<br />
k=0<br />
( n−1<br />
)<br />
k Pk (0) t (n−1)−k<br />
= nP n−1 (t) if n = 1, 2, ...<br />
(e) P n (t + 1) − P n (t) = nt n−1 if n = 1, 2, ...<br />
37
Proof: Consider<br />
f t+1 (x) − f t (x) =<br />
so as n = 1, 2, ..., we have<br />
∞∑<br />
n=0<br />
[P n (t + 1) − P n (t)] xn<br />
n!<br />
= xe xt by f t (x) = xe xt / (e x − 1)<br />
∞∑<br />
= (n + 1) t n xn+1<br />
(n + 1)! ,<br />
n=0<br />
P n (t + 1) − P n (t) = nt n−1 .<br />
by (b)<br />
(f) P n (1 − t) = (−1) n P n (t)<br />
Proof: Note that<br />
so we have<br />
n=0<br />
Hence, P n (1 − t) = (−1) n P n (t) .<br />
(g) B 2n+1 = 0 if n = 1, 2, ...<br />
f t (−x) = f 1−t (x) ,<br />
∞∑<br />
∞<br />
(−1) n P n (t) xn<br />
n! = ∑<br />
P n (1 − t) xn<br />
n! .<br />
Proof: With help of (e) and (f), let t = 0 and n = 2k + 1, then it is clear<br />
that B 2k+1 = 0 if k = 1, 2, ...<br />
n=0<br />
(h) 1 n + 2 n + ... + (k − 1) n = P n+1(k)−P n+1 (0)<br />
n+1<br />
(n = 2, 3, ...)<br />
Proof: With help of (e), we know that<br />
P n+1 (t + 1) − P n+1 (t)<br />
n + 1<br />
= t n<br />
which implies that<br />
1 n + 2 n + ... + (k − 1) n = P n+1 (k) − P n+1 (0)<br />
n + 1<br />
(n = 2, 3, ...)<br />
38
Remark: (1) <strong>The</strong> reader can see the book, Infinite Series by Chao<br />
Wen-Min, pp 355-366. (Chinese Version)<br />
(2) <strong>The</strong>re are some special polynomials worth studying, such as Legengre<br />
Polynomials. <strong>The</strong> reader can see the book, Essentials of Ordinary<br />
Differential Equations by Ravi P. Agarwal and Ramesh C. Gupta.<br />
pp 305-312.<br />
(3) <strong>The</strong> part (h) tells us one formula to calculte the value of the finite<br />
series ∑ m<br />
k=1 kn . <strong>The</strong>re is an interesting story from the mail that Fermat,<br />
pierre de (1601-1665) sent to Blaise Pascal (1623-1662). Fermat<br />
used the Mathematical Induction to show that<br />
n∑<br />
k (k + 1) · · · (k + p) =<br />
k=1<br />
n (n + 1) · · · (n + p + 1)<br />
. (*)<br />
p + 2<br />
In terms of (*), we can obtain another formula on ∑ m<br />
k=1 kn .<br />
39
Limit sup and limit inf.<br />
Introduction<br />
In order to make us understand the information more on approaches of a given real<br />
sequence a n <br />
n1<br />
, we give two definitions, thier names are upper limit and lower limit. It<br />
is fundamental but important tools in analysis.<br />
Definition<br />
Definition of limit sup and limit inf<br />
and<br />
Example<br />
Example<br />
Example<br />
Given a real sequence a n <br />
n1<br />
,wedefine<br />
b n supa m : m n<br />
1 1 n <br />
n1<br />
1 n <br />
n n1<br />
<br />
n n1<br />
c n infa m : m n.<br />
0,2,0,2,..., sowehave<br />
b n 2andc n 0 for all n.<br />
1, 2, 3,4,..., sowehave<br />
b n and c n for all n.<br />
1, 2, 3, . . . , sowehave<br />
b n n and c n for all n.<br />
Proposition Given a real sequence a n <br />
n1<br />
, and thus define b n and c n as the same as<br />
before.<br />
1 b n ,andc n n N.<br />
2 If there is a positive integer p such that b p , then b n n N.<br />
If there is a positive integer q such that c q , then c n n N.<br />
3 b n is decreasing and c n is increasing.<br />
By property 3, we can give definitions on the upper limit and the lower limit of a given<br />
sequence as follows.<br />
Definition Given a real sequence a n and let b n and c n as the same as before.<br />
(1) If every b n R, then<br />
infb n : n N<br />
is called the upper limit of a n , denoted by<br />
lim n<br />
sup a n .<br />
That is,<br />
lim n<br />
sup a n inf b n.<br />
n<br />
If every b n , then we define<br />
lim n<br />
sup a n .<br />
(2) If every c n R, then<br />
supc n : n N<br />
is called the lower limit of a n , denoted by<br />
lim n<br />
inf a n .
That is,<br />
lim n<br />
inf a n sup c n .<br />
n<br />
If every c n , then we define<br />
lim n<br />
inf a n .<br />
Remark <strong>The</strong> concept of lower limit and upper limit first appear in the book (Analyse<br />
Alge’brique) written by Cauchy in 1821. But until 1882, Paul du Bois-Reymond<br />
gave explanations on them, it becomes well-known.<br />
Example 1 1 n <br />
n1<br />
0,2,0,2,..., sowehave<br />
b n 2andc n 0 for all n<br />
which implies that<br />
lim sup a n 2 and lim inf a n 0.<br />
Example 1 n <br />
n n1<br />
1, 2, 3,4,..., sowehave<br />
b n and c n for all n<br />
which implies that<br />
lim sup a n and lim inf a n .<br />
<br />
n n1<br />
Example 1, 2, 3, . . . , sowehave<br />
b n n and c n for all n<br />
which implies that<br />
lim sup a n and lim inf a n .<br />
Relations with convergence and divergence for upper (lower) limit<br />
<strong>The</strong>orem Let a n be a real sequence, then a n converges if, and only if, the upper<br />
limit and the lower limit are real with<br />
lim n<br />
sup a n lim n<br />
inf a n lim n<br />
a n .<br />
<strong>The</strong>orem Let a n be a real sequence, then we have<br />
(1) lim n sup a n a n has no upper bound.<br />
(2) lim n sup a n for any M 0, there is a positive integer n 0 such<br />
that as n n 0 ,wehave<br />
a n M.<br />
(3) lim n sup a n a if, and only if, (a) given any 0, there are infinite<br />
many numbers n such that<br />
a a n<br />
and (b) given any 0, there is a positive integer n 0 such that as n n 0 ,wehave<br />
a n a .<br />
Similarly, we also have<br />
<strong>The</strong>orem Let a n be a real sequence, then we have<br />
(1) lim n inf a n a n has no lower bound.<br />
(2) lim n inf a n for any M 0, there is a positive integer n 0 such
that as n n 0 ,wehave<br />
a n M.<br />
(3) lim n inf a n a if, and only if, (a) given any 0, there are infinite<br />
many numbers n such that<br />
a a n<br />
and (b) given any 0, there is a positive integer n 0 such that as n n 0 ,wehave<br />
a n a .<br />
From <strong>The</strong>orem 2 an <strong>The</strong>orem 3, the sequence is divergent, we give the following<br />
definitios.<br />
Definition Let a n be a real sequence, then we have<br />
(1) If lim n sup a n , then we call the sequence a n diverges to ,<br />
denoted by<br />
lim n<br />
a n .<br />
<strong>The</strong>orem<br />
<strong>The</strong>orem<br />
<strong>The</strong>orem<br />
(2) If lim n inf a n , then we call the sequence a n diverges to ,<br />
denoted by<br />
lim n<br />
a n .<br />
Let a n be a real sequence. If a is a limit point of a n , then we have<br />
lim n<br />
inf a n a lim n<br />
sup a n .<br />
Some useful results<br />
Let a n be a real sequence, then<br />
(1) lim n inf a n lim n sup a n .<br />
(2) lim n infa n lim n sup a n and lim n supa n lim n inf a n<br />
(3) If every a n 0, and 0 lim n inf a n lim n sup a n , then we<br />
have<br />
lim n<br />
sup a 1 n<br />
1<br />
lim n inf a n<br />
Let a n and b n be two real sequences.<br />
(1) If there is a positive integer n 0 such that a n b n , then we have<br />
lim n<br />
inf a n lim n<br />
inf b n and lim n<br />
sup a n lim n<br />
sup b n .<br />
(2) Suppose that lim n inf a n , lim n inf b n , lim n sup a n ,<br />
lim n sup b n , then<br />
lim n<br />
inf a n lim n<br />
inf b n<br />
and lim n<br />
inf 1 a n<br />
1<br />
lim n sup a n<br />
.<br />
lim n<br />
infa n b n <br />
lim n<br />
inf a n lim n<br />
sup b n (or lim n<br />
sup a n lim n<br />
inf b n )<br />
lim n<br />
supa n b n <br />
lim n<br />
sup a n lim n<br />
sup b n .<br />
In particular, if a n converges, we have<br />
lim n<br />
supa n b n lim n<br />
a n lim n<br />
sup b n
and<br />
lim n<br />
infa n b n lim n<br />
a n lim n<br />
inf b n .<br />
(3) Suppose that lim n inf a n , lim n inf b n , lim n sup a n ,<br />
lim n sup b n , anda n 0, b n 0 n, then<br />
lim n<br />
inf a n<br />
lim n<br />
inf b n<br />
lim n<br />
infa n b n <br />
lim n<br />
inf a n lim n<br />
sup b n (or lim n<br />
inf b n<br />
lim n<br />
supa n b n <br />
lim n<br />
sup a n lim n<br />
sup b n .<br />
In particular, if a n converges, we have<br />
and<br />
lim n<br />
supa n b n <br />
lim n<br />
infa n b n <br />
lim n<br />
a n<br />
lim n<br />
a n<br />
lim n<br />
sup b n<br />
lim n<br />
inf b n .<br />
lim n<br />
sup a n )<br />
<strong>The</strong>orem Let a n be a positive real sequence, then<br />
lim n<br />
inf a n1<br />
a n<br />
lim n<br />
infa n 1/n lim n<br />
supa n 1/n lim n<br />
sup a n1<br />
a n<br />
.<br />
Remark We can use the inequalities to show<br />
n! 1/n<br />
lim n n 1/e.<br />
<strong>The</strong>orem Let a n be a real sequence, then<br />
lim n<br />
inf a n lim n<br />
inf a 1 ...a n<br />
n<br />
lim n<br />
sup a 1 ...a n<br />
n<br />
lim n<br />
sup a n .<br />
Exercise Let f : a, d R be a continuous function, and a n is a real sequence. If f is<br />
increasing and for every n, lim n inf a n , lim n sup a n a, d, then<br />
lim n<br />
sup fa n f lim n<br />
sup a n<br />
and lim n<br />
inf fa n f lim n<br />
inf a n .<br />
Remark: (1) <strong>The</strong> condition that f is increasing cannot be removed. For<br />
example,<br />
fx |x|,<br />
and<br />
1/k if k is even<br />
a k <br />
1 1/k if k is odd.<br />
(2) <strong>The</strong> proof is easy if we list the definition of limit sup and limit inf. So, we<br />
omit it.<br />
Exercise Let a n be a real sequence satisfying a np a n a p for all n, p. Show that<br />
an<br />
n converges.<br />
Hint: Consider its limit inf.
Something around the number e<br />
1. Show that the sequence 1 1 n n converges, and denote the limit by e.<br />
Proof: Since<br />
1 1 n<br />
n<br />
n<br />
kn 1 n<br />
k<br />
k0<br />
1 n 1 n<br />
<br />
nn 1<br />
2!<br />
1<br />
n<br />
1 1 <br />
2!<br />
1 1 1 n ...<br />
n!<br />
1<br />
1 1 1 2 1 2 .. 1 ...<br />
2 2 n1<br />
3,<br />
2 nn 1 1<br />
.. 1<br />
n! n<br />
1 1 n 1 n n<br />
1<br />
and by (1), we know that the sequence is increasing. Hence, the sequence is convergent.<br />
We denote its limit e. Thatis,<br />
lim n<br />
1 1 n<br />
n e.<br />
n<br />
1<br />
Remark: 1. <strong>The</strong> sequence and e first appear in the mail that Euler wrote to Goldbach.<br />
It is a beautiful formula involving<br />
e i 1 0.<br />
1<br />
2. Use the exercise, we can show that k0<br />
e as follows.<br />
k!<br />
Proof: Let x n 1 1 n n , and let k n, wehave<br />
1 1 <br />
2!<br />
1 1 1 ..<br />
k n!<br />
1 1 1 1 <br />
k<br />
n k<br />
1 x k<br />
which implies that ( let k )<br />
On the other hand,<br />
So,by(2)and(3),wefinallyhave<br />
n<br />
y n : <br />
i0<br />
<br />
<br />
k0<br />
1<br />
i! e. 2<br />
x n y n 3<br />
1<br />
k! e. 4<br />
3. e is an irrational number.<br />
Proof: Assume that e is a rational number, say e p/q, where g.c.d. p, q 1. Note<br />
that q 1. Consider<br />
q!e q!<br />
q!<br />
<br />
<br />
k0<br />
q<br />
<br />
k0<br />
1<br />
k!<br />
1<br />
k!<br />
q!<br />
<br />
<br />
kq1<br />
and since q! k0<br />
q 1<br />
k!<br />
and q!e are integers, we have q! kq1<br />
1<br />
k!<br />
is also an<br />
integer. However,<br />
1<br />
k!<br />
,
q!<br />
<br />
<br />
kq1<br />
1<br />
k!<br />
<br />
<br />
kq1<br />
q!<br />
k!<br />
1<br />
q 1 1<br />
q 1q 2 ...<br />
1<br />
q 1 1<br />
q 1<br />
2<br />
...<br />
1 q<br />
1,<br />
a contradiction. So, we know that e is not a rational number.<br />
n 1<br />
4. Here is an estimate about e k0 k!<br />
, where 0 1. ( In fact, we know<br />
nn!<br />
that e 2. 71828 18284 59045 . . . . )<br />
<br />
Proof: Since e k0<br />
1<br />
k!<br />
<br />
0 e x n <br />
kn1<br />
1<br />
n 1!<br />
1<br />
n 1!<br />
1<br />
k! ,wherex n <br />
n<br />
k0<br />
1<br />
k!<br />
1 1<br />
n 2 1<br />
n 2n 3 ...<br />
1 1<br />
n 2 1<br />
n 2 2<br />
1<br />
n 1! n n 2<br />
1<br />
1<br />
nn! since n 2 1<br />
n 1 2 n .<br />
So, we finally have<br />
n<br />
e <br />
1<br />
k! , where 0 1.<br />
nn!<br />
k0<br />
Note: We can use the estimate dorectly to show e is an irrational number.<br />
2. For continuous variables, we have the samae result as follows. That is,<br />
x<br />
e.<br />
lim<br />
x<br />
1 1 x<br />
Proof: (1) Since 1 1 n n e as n , we know that for any sequence a n N,<br />
with a n , wehave<br />
lim n<br />
1 a 1 a n<br />
n<br />
e. 5<br />
(2) Given a sequence x n with x n , and define a n x n , then<br />
a n x n a n 1, then we have<br />
a n x<br />
1 1 1 1 n<br />
a n 1<br />
xn<br />
1 a 1 a n1<br />
n<br />
.<br />
Since<br />
a n a<br />
1 1 e and 1 1 n1<br />
a n 1<br />
an e as x by (5)<br />
we know that<br />
...
lim 1 1 x n<br />
n x n<br />
e.<br />
Since x n is arbitrary chosen so that it goes infinity, we finally obtain that<br />
lim 1 1 x<br />
x x e. 6<br />
(3) In order to show 1 1 x x e as x , weletx y, then<br />
1 1 x y<br />
x 1 1 y<br />
<br />
y<br />
y 1<br />
1 1 1 1<br />
y 1 y 1 .<br />
Note that x y , by (6), we have shown that<br />
e y<br />
lim 1 1<br />
y 1<br />
lim x<br />
1 1 x<br />
x<br />
.<br />
y<br />
y1<br />
y1<br />
1 1<br />
y 1<br />
3. Prove that as x 0, we have1 1 x x is strictly increasing, and 1 1 x x1 is<br />
dstrictly ecreasing.<br />
Proof: Since, by Mean Value <strong>The</strong>orem<br />
1<br />
x 1 log 1 1 x logx 1 logx 1 1 x for all x 0,<br />
we have<br />
x log 1 1 <br />
x log 1 1 x 1 0 for all x 0<br />
x 1<br />
and<br />
<br />
x 1 log 1 1 x log 1 1 x 1 x 0 for all x 0.<br />
Hence, we know that<br />
x log 1 1 x is strictly increasing on 0, <br />
and<br />
x 1 log 1 1 x is strictly decreasing on 0, .<br />
It implies that<br />
1 1 x<br />
x<br />
is strictly increasing 0, , and 1 1 x<br />
x1<br />
is strictly decreasing on 0, .<br />
Remark: By exercise 2, we know that<br />
lim<br />
x<br />
1 1 x<br />
x<br />
e lim<br />
x<br />
1 1 x<br />
4. Follow the Exercise 3 to find the smallest a such that 1 1 x xa e and strictly<br />
decreasing for all x 0, .<br />
Proof: Let fx 1 1 x xa , and consider<br />
log fx x a log 1 1 x : gx,<br />
Let us consider<br />
x1<br />
.
g x log 1 1 x x a<br />
x 2 x<br />
log1 y y 1 ay 2 1<br />
1 y , where 0 y 1<br />
1 x 1<br />
<br />
<br />
y<br />
k<br />
k y 1 ay2 y k<br />
k1<br />
k0<br />
1<br />
2<br />
a y 2 1<br />
3<br />
a y 3 ... 1 n a yn ...<br />
It is clear that for a 1/2, we have g x 0 for all x 0, . Note that for a 1/2,<br />
if there exists such a so that f is strictly decreasing for all x 0, . <strong>The</strong>n g x 0for<br />
all x 0, . However, it is impossible since<br />
g x 1 a y 2 1 a y 3 ... 1 2 3 n a yn ...<br />
1 2 a 0asy 1 .<br />
So, we have proved that the smallest value of a is 1/2.<br />
Remark: <strong>The</strong>re is another proof to show that 1 1 x x1/2 is strictly decreasing on<br />
0, .<br />
Proof: Consider ht 1/t, and two points 1, 1 and 1 1 1<br />
x , lying on the graph<br />
1 1 x<br />
From three areas, the idea is that<br />
<strong>The</strong> area of lower rectangle <strong>The</strong>areaofthecurve <strong>The</strong>areaoftrapezoid<br />
So, we have<br />
1<br />
1 x 1 x<br />
Consider<br />
1 1 x<br />
1<br />
1 1 x<br />
x1/2<br />
<br />
<br />
log 1 1 x 1 2x<br />
1 1 x<br />
x1/2<br />
1 1<br />
1 1 x<br />
log 1 1 x x 1 2<br />
x 1 2<br />
1<br />
xx 1<br />
1<br />
xx 1<br />
0by(7);<br />
hence, we know that 1 1 x x1/2 is strictly decreasing on 0, .<br />
Note: Use the method of remark, we know that 1 1 x x is strictly increasing on<br />
0, .<br />
. 7